Question

007 10.0 points
a block is released from rest, at a height h,
and allowed to slide down an inclined plane.
there is friction on the plane. at the bottom
of the plane, there is a spring that the block
will compress. after compressing the spring,
the block will slide up the plane to some maxi-
mum height, ha, after which it will again slide
back down.
how much work is done on the block be-
tween its release at height h and its ascent to
its next maximum height?

1. 0
2. none of these
3. more information is needed.
4. 2μ mag ha
5. mag (h - ha)
6. mag (h - ha) + 2μ mag ha

Explanation:

Step1: Apply work - energy theorem

The work - energy theorem states that \(W_{net}=\Delta K\). The initial kinetic energy \(K_{i} = 0\) (released from rest) and the final kinetic energy \(K_{f}=0\) (at maximum height \(h_{A}\)). The total work done on the block is the sum of the work done by gravity \(W_{g}\) and the work done by friction \(W_{f}\).

The work done by gravity is \(W_{g}=mg(h - h_{A})\) (using \(W = Fd\cos\theta\), where for gravity \(F = mg\), \(d=(h - h_{A})\) and \(\theta = 0\), \(\cos\theta=1\)).

Let the length of the inclined - plane be \(l\). The normal force on the inclined - plane \(N = mg\cos\theta\). The force of friction \(f=\mu N=\mu mg\cos\theta\). When the block slides down and then up, the distance traveled due to friction is \(2l\). Also, using trigonometry \(h = l\sin\theta\) and \(h_{A}=l_{A}\sin\theta\). The work done by friction \(W_{f}=- 2\mu mgh_{A}\cot\theta\) (the negative sign because friction is opposite to the direction of motion). But if we consider the energy approach in terms of height differences:

The total work done \(W = W_{g}+W_{f}\). Since \(K_{i} = K_{f} = 0\), from \(W=\Delta K\), we know that the work done by non - conservative forces (friction) and conservative forces (gravity) together is zero.

The work done by gravity is \(W_{g}=mg(h - h_{A})\) (positive as the block moves down relative to the initial height and up to a lower height). The work done by friction is \(W_{f}=-mg(h - h_{A})\) (because the net change in kinetic energy is zero).

Another way: Using the formula \(W=\Delta U+\Delta K\). \(\Delta K = 0\), \(\Delta U=mg(h_{A}-h)\). The work done by non - conservative forces (friction) \(W_{nc}\):

\(W_{nc}+W_{c}=\Delta K\). \(W_{c}=- \Delta U\) (work done by conservative force \(W_{c}=mg(h - h_{A})\)), and \(W_{nc}\) (work done by friction) cancels out \(W_{c}\) since \(\Delta K = 0\).

If we consider the work - energy theorem for the whole process (from height \(h\) to height \(h_{A}\)):

The work done by all forces \(W=\Delta K\). The forces acting are gravity (\(F_{g}=mg\)) and friction (\(f\)).

The displacement of the block (in terms of height) is \((h - h_{A})\). The work done by gravity \(W_{g}=mg(h - h_{A})\) (positive as the block moves in the direction of the gravitational force component along the displacement in terms of height).

Since the block starts and ends at rest (\(K_{i}=K_{f} = 0\)), the work done by friction \(W_{f}=-mg(h - h_{A})\). But if we consider the work - energy theorem in terms of the path:

Let the length of the incline be \(L\) for the first (downward) motion and \(L_{A}\) for the upward motion. \(h = L\sin\theta\), \(h_{A}=L_{A}\sin\theta\). The work done by gravity \(W_{g}=mg(h - h_{A})\), the work done by friction \(W_{f}=-\mu mg\cos\theta(L + L_{A})\). Also, using \(L=\frac{h}{\sin\theta}\) and \(L_{A}=\frac{h_{A}}{\sin\theta}\), \(W_{f}=-\mu mg\cot\theta(h + h_{A})\). But from \(K_{i} = K_{f}=0\), \(W_{g}+W_{f}=0\)

In terms of the given options, if we assume the problem is using the fact that the work done by all forces (gravity and friction) is zero (since \(\Delta K = 0\)). The work done by gravity is \(mg(h - h_{A})\) (positive) and the work done by friction is \(-mg(h - h_{A})\).

Answer:

1. \(0\)