Question

A package weighing 2000 N is placed on a ramp and attached to a cable that runs over a massless, frictionless pulley, and a force F is applied to the cable. The ramp is inclined at an angle of 42° with respect to the horizontal, and the coefficient of kinetic friction is 0.2. Determine the minimum force F needed to prevent the block from sliding down the ramp.

A. 1200 N
B. 1800 N
C. 2000 N
D. 2400 N

(b) The cable suddenly snaps. What is the acceleration of the block?
A. 9.8 m/s²
B. 5.7 m/s²
C. 4.9 m/s²
D. 2.4 m/s²

Answer 1

To find the minimum force F to prevent the block from sliding down the ramp, one must consider the gravitational force, the kinetic frictional force, and the tension in the cable. The force F is calculated based on the sum of these forces. When the cable snaps, the acceleration of the block can be determined by subtracting the frictional force from the gravitational force component along the ramp.

Step-by-step explanation:

To determine the minimum force F needed to prevent the block from sliding down the ramp, we must consider the component of gravitational force parallel to the ramp, the force of kinetic friction, and the force exerted by the cable. The gravitational force parallel to the ramp is Wsin(θ) = 2000 N sin(42°), and the frictional force is f_k = μkN = 0.2 × (2000 N cos(42°)). The sum of these two forces equals the force F needed to prevent sliding. Solving for F gives us the answer.

If the cable snaps, the block will start to accelerate down the ramp due to the net force acting on it, which is the component of gravity minus friction. This acceleration can be calculated using Newton's second law, a = (W sin(θ) - f_k)/m, where m is the mass of the package (2000 N divided by 9.8 m/s²).