AP Physics C Mechanics - 3.2 Work- Exam Style questions- MCQs
Work AP Physics C Mechanics MCQ
Unit 3: Work, Energy, and Power
Weightage : 15-25%
Unit 3: Work, Energy, and Power
Weightage : 15-25%
A block of mass \(m\) experiences a force \(F\), which pushes downward and to the right at an angle \(\theta\), as shown above. The block moves across a rough horizontal surface with coefficient of kinetic friction \(\mu\).
As the block moves a horizontal distance \(d\), how much work is done by the net force on the block?
Correct Answer: \( \boxed{\mathrm{E}} \)
Only the horizontal forces do work because the displacement is entirely horizontal.
The horizontal component of the applied force is
\(F_x=F\cos\theta\)
The downward component of the applied force increases the normal force:
\(N=mg+F\sin\theta\)
Therefore, the kinetic friction force is
\(f_k=\mu N=\mu\left(mg+F\sin\theta\right)\)
The net horizontal force is
\(F_{\mathrm{net}}=F\cos\theta-\mu\left(mg+F\sin\theta\right)\)
Since the net force is parallel to the displacement,
\(W_{\mathrm{net}}=F_{\mathrm{net}}d\)
Hence,
\(W_{\mathrm{net}}=\left(F\cos\theta-\mu\left(mg+F\sin\theta\right)\right)d\)
Therefore, the correct answer is (E).
A motorcycle of mass \(200\,\mathrm{kg}\) completes a vertical circular loop of radius \(5\,\mathrm{m}\) with a constant speed of \(10\,\mathrm{m/s}\).
How much work is done on the motorcycle by the normal force of the track?
Correct Answer: \( \boxed{\mathrm{A}} \)
Work done by a force is given by
\(W=\vec{F}\cdot\vec{s}=Fs\cos\theta\)
At every point on the circular path, the normal force acts radially toward the center of the loop, while the motorcycle’s instantaneous displacement is tangent to the path.
Therefore, the angle between the normal force and the displacement is
\(\theta=90^\circ\)
Since \(\cos90^\circ=0\),
\(W_{\mathrm{N}}=Fs\cos90^\circ=0\)
Thus, the normal force changes only the direction of the motorcycle’s velocity and does not change its kinetic energy.
Therefore, the work done by the normal force is \(0\,\mathrm{J}\).
Hence, the correct answer is (A).
A student pushes a box across a rough horizontal floor. If the amount of work done by the student on the box is \(100\,\mathrm{J}\) and the amount of energy dissipated by friction is \(40\,\mathrm{J}\), what is the change in kinetic energy of the box?
Correct Answer: \( \boxed{\mathrm{C}} \)
According to the Work-Energy Theorem, the change in kinetic energy equals the net work done on the box:
\(\Delta K=W_{\mathrm{net}}\)
The student does positive work of
\(W_{\mathrm{student}}=100\,\mathrm{J}\)
Friction does negative work of
\(W_{\mathrm{friction}}=-40\,\mathrm{J}\)
Therefore, the net work is
\(W_{\mathrm{net}}=100-40=60\,\mathrm{J}\)
Hence, the change in the box’s kinetic energy is
\(\Delta K=60\,\mathrm{J}\)
Therefore, the correct answer is (C).