Unit 2: Force and Translational Dynamics
Weightage : 20-15%
A block of mass \(m\) is initially at rest on a rough horizontal surface when a constant force \(F\) is exerted on it, as shown above. The block accelerates to the right and is moving with speed \(v\) after it has traveled a distance \(d\).
Which of the following equations can be used to determine the frictional force exerted on the block by the surface?
Correct Answer: \( \boxed{\mathrm{C}} \)
Since the block starts from rest, use the kinematic equation
\(v^2=v_0^2+2ad\)
With \(v_0=0\),
\(a=\dfrac{v^2}{2d}\)
Applying Newton’s second law in the horizontal direction,
\(\sum F=ma\)
The applied force acts to the right and friction acts to the left, so
\(F-f=ma\)
Substituting the expression for the acceleration,
\(F-f=m\left(\dfrac{v^2}{2d}\right)=\dfrac{mv^2}{2d}\)
This equation can be rearranged to solve for the frictional force if desired.
Therefore, the correct answer is (C).
A block of mass \(m\) is pushed up a rough inclined plane by a force of magnitude \(F\), as shown above. The frictional force between the block and the surface has magnitude \(f\). The plane makes an angle \(\theta\) with the horizontal.
Which of the following equations can be used to determine the acceleration of the block?
Correct Answer: \( \boxed{\mathrm{C}} \)
Choose the positive direction to be up the incline.
The applied force \(F\) acts up the incline, while both the frictional force \(f\) and the component of gravity parallel to the incline, \(mg\sin\theta\), act down the incline.
Applying Newton’s second law along the incline,
\(\sum F=ma\)
gives
\(F-f-mg\sin\theta=ma\)
The perpendicular component of gravity, \(mg\cos\theta\), is balanced by the normal force and therefore does not contribute to the acceleration along the incline.
Therefore, the correct answer is (C).
Blocks A and B of unknown masses \(m_1\) and \(m_2\), respectively, are arranged as shown above. Block A rests on a frictionless incline and is connected to block B by a light string passing over an ideal pulley. The blocks are released from rest, and block A accelerates up the incline with measured acceleration \(a\).
Which of the following correctly relates the measured acceleration to the masses of the blocks?
Correct Answer: \( \boxed{\mathrm{E}} \)
Apply Newton’s second law to the two-block system along the direction of motion:
\(m_2g-m_1g\sin\theta=(m_1+m_2)a\)
Solving for the acceleration gives
\(a=\dfrac{m_2g-m_1g\sin\theta}{m_1+m_2}\)
The measured acceleration depends on both masses simultaneously. It compares the quantities \(m_2g\) and \(m_1g\sin\theta\), not the masses \(m_1\) and \(m_2\) directly.
For different values of \(\theta\), two different mass combinations can produce the same acceleration. Therefore, knowing only the acceleration is insufficient to determine whether \(m_1\) is greater than \(m_2\).
Therefore, the correct answer is (E).
