Worked solutions

The actual forces are the pull, weight down and normal reaction up.

Vertical equilibrium gives N + 10 − 40 = 0, so N = 30 N.

Check: N + Fy = mg. The horizontal and vertical components resolve the one applied force.

Since the block slides right, friction points left:

The pull partly supports the block, so using μkmg would overestimate friction. The positive acceleration means the block speeds up in its current rightward motion.

During rest, N = 40 − F/2 and the required leftward static friction is F cos30°.

Rest is possible while:

The limiting force is:

At exactly this value the block is at limiting equilibrium; motion begins when the applied force exceeds it.

For F = 12 N, N = 34 N. The required static friction is 6√3 ≈ 10.3923 N, less than its 17 N maximum. Its actual value is 10.3923 N leftward.

For F = 20 N, the required 17.3205 N exceeds the 15 N static maximum. Rightward sliding begins. Kinetic friction then becomes 7.50 N, giving ax ≈ 2.4551 m/s².

N first reaches zero at F = 80 N. At this boundary the surface is unloaded; for larger F the block accelerates upward and the contact equations must change.

Check: μsN is a maximum, not a value to assign at every force.

At the threshold:

The denominator can be written:

It is largest at θ = tan⁻¹μs, so:

For rightward sliding:

Its maximum is at θ = tan⁻¹μk ≈ 14.0362°:

Both have N > 0. The first optimum concerns limiting static friction; the second concerns kinetic friction during an existing motion.

Check: the best angle is not necessarily the angle giving the largest horizontal component alone.

At the contact boundary:

At 60° the upward component exceeds the weight. The block leaves the surface, so N = 0:

A negative normal reaction would require the floor to attract the block. It signals that the assumed contact model has ceased to apply.

The floor reaction is (1 + 3)g = 40 N, so floor friction is 8 N leftward.

If both blocks move together:

For F = 24 N, a = 4 m/s². The upper block requires 4 N of rightward friction. The equal and opposite contact friction on the lower block is leftward.

The upper block’s maximum static friction is 0.50 × 1 × 10 = 5 N. Hence a ≤ 5 m/s²:

With F = 32 N, relative sliding begins. Interface kinetic friction has magnitude 0.30 × 10 = 3 N:

The lower block accelerates faster, consistent with the assumed relative slipping direction.

Initially the spring pulls right with k(0.40) = 20 N, exceeding the 3 N static maximum. Motion begins.

While the block moves right, kinetic friction is 2 N leftward. At its first turning point x > 0:

Thus 25x² + 2x − 3.2 = 0. The physically relevant root is x = +0.32 m.

During the first rightward motion:

Speed is largest where the rightward net force vanishes:

The kinetic energy there is 3.24 J, so:

At x = +0.32 m, the spring pulls left with 16 N. Static friction can provide at most 3 N, so the block reverses.

For successive turning amplitudes A and B, energy loss over distance A + B gives:

so A − B = 2(2)/50 = 0.08 m. The turning positions are:

−0.40, +0.32, −0.24, +0.16, −0.08, 0 m.

At every nonzero turning point the spring exceeds the 3 N static limit. At x = 0 it has no force and remains at rest.

Check: maximum speed occurs before the relaxed position because friction also acts there.

Crossing the rough section dissipates μkmgL. The loop entry speed satisfies:

At the top, energy gives vt² = vb² − 4gR. Contact requires vt² ≥ gR. Therefore:

At that height vb² = 5gR. The bottom reaction is:

For a general angle θ from the bottom:

For H = 1.20 m, set N = 0:

The speed is still positive:

It therefore loses contact before reversing on the track.

Correct statements: A, B and D.

Check: energy determines the available speed; the normal reaction determines whether the prescribed circular path is physically possible.

(a) (x − 3)(x − 4) ≤ 0 gives [3, 4].

(b) (x + 3)(x − 2) > 0 gives (−∞, −3) ∪ (2, ∞).

(c) Multiplying by −1 reverses the inequality. Then (x − 1)(x − 3) ≥ 0 gives (−∞, 1] ∪ [3, ∞).

Check one point inside and outside each pair of roots. Use brackets only where equality is permitted.

The domain excludes x = −1 and x = 2.

The squared factor is positive except at x = 1, where it is zero. Away from that point, the sign follows:

The signs on the intervals separated by −1, 1, 2 and 4 are:

Interval Sign

x < −1 Negative

−1 < x < 1 Positive

1 < x < 2 Positive

2 < x < 4 Negative

x > 4 Positive

Include x = 1 as an isolated zero and x = 4 as an allowed endpoint. The answer is:

Check: cancelling the squared factor without separately keeping x = 1 would lose a valid solution.

If a < 1, the quadratic opens downward and becomes negative for large |x|.

If a = 1, the expression is −2x + 4, which also fails for all real x.

For a > 1, require a nonpositive discriminant:

Thus:

At a = 3/2 the expression is ½(x − 3)², so equality is valid.

Check: the discriminant condition was used only after verifying a positive leading coefficient.

The right side must satisfy x + 1 ≥ 0.

The absolute value condition is equivalent to:

and:

The second expression is always positive because it opens upward and has discriminant −7. The first is nonpositive between its roots:

This interval already has x > −1, so the right side condition is satisfied.

Check: squaring without checking signs can introduce extraneous solutions.

For a convex quadratic to have two distinct roots inside (0, 3), its minimum must be inside that interval and below zero, with positive values at both endpoints.

The vertex is x = k. The discriminant is:

Together with 0 < k < 3 this gives k > 2.

At x = 0 the value k + 2 is positive. At x = 3:

Therefore:

At k = 2 the roots merge. At k = 11/5 one root is 3. Both boundaries must be excluded.

The domain is x > 1 with x ≠ 2. The argument is always positive:

For 1 < x < 2 the base is between 0 and 1, so:

Its solution [2, 4] has no overlap with (1, 2).

For x > 2, the base exceeds 1:

Intersecting with x > 2 gives:

Check: at x = 4 the logarithm is log₃3 = 1.

For a ≤ 0, the vertex is at or left of the interval. The minimum on [0, 2] is at x = 0 and equals a + 1. These values work exactly when −1 ≤ a ≤ 0.

For 0 < a < 2, the vertex lies inside the interval. Its value is:

It is nonnegative for:

For a ≥ 2, the minimum occurs at x = 2, where 5 − 3a < 0. These values fail.

Therefore:

At a = −1 the expression is x(x + 2), which is nonnegative on [0, 2] even though it has real roots and takes negative values outside that interval. At the upper boundary it is a perfect square.

Requiring a nonpositive discriminant would incorrectly exclude −1 ≤ a < (1 − √5)/2. Roots and negative values outside the requested interval need not invalidate the condition.

The derivative is:

For a < 0 it is positive everywhere. For a = 0 it vanishes only at x = 0 and the cubic remains strictly increasing. For a > 0 it is negative between −√a and √a.

Answer to (a): a ≤ 0.

For a > 0 the stationary values are:

Three distinct roots require the local maximum to be positive and the local minimum negative. Hence:

To put all three inside (−3, 3), require the two stationary points inside the interval, f(−3) < 0 and f(3) > 0.

The stricter bound is a < 25/9. With a > 1 this also places the stationary points inside (−3, 3). Thus:

At a = 1:

There are two distinct real roots, one repeated. At a = 25/9 one root reaches −3.

Check: a derivative equal to zero at one point does not by itself prevent strict increase or guarantee an extremum.