Question

A contestant in a winter games event pushes a 59.0 kg block of ice across a frozen lake as shown in the figure.

The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
(a)
Calculate the minimum force F (in N) he must exert to get the block moving.

(b)
What is its acceleration (in m/s2) once it starts to move, if that force is maintained?

Answer 1

(a) The minimum force required to get the block moving can be calculated using the formula F = μ * m * g, where μ is the coefficient of static friction, m is the mass of the block, and g is the acceleration due to gravity. Plugging in the values given in the problem, we get F = 0.1 * 59.0 kg * 9.8 m/s² = 57.9 N. So the minimum force required to get the block moving is 57.9 N.

(b) Once the block starts to move, its acceleration can be calculated using the formula a = (F - f_k) / m, where F is the applied force, f_k is the force of kinetic friction, and m is the mass of the block. The force of kinetic friction can be calculated using the formula f_k = μ_k * m * g, where μ_k is the coefficient of kinetic friction. Plugging in the values given in the problem, we get f_k = 0.03 * 59.0 kg * 9.8 m/s² = 17.4 N. So if the applied force of 57.9 N is maintained, then the acceleration of the block would be a = (57.9 N - 17.4 N) / 59.0 kg = 0.69 m/s².

Answer 2

(a) The minimum force required to get the block moving can be calculated using the formula F = μ * m * g, where μ is the coefficient of static friction, m is the mass of the block, and g is the acceleration due to gravity. Plugging in the values given in the problem, we get F = 0.1 * 59.0 kg * 9.8 m/s² = 57.9 N. So the minimum force required to get the block moving is 57.9 N.

(b) Once the block starts to move, its acceleration can be calculated using the formula a = (F - f_k) / m, where F is the applied force, f_k is the force of kinetic friction, and m is the mass of the block. The force of kinetic friction can be calculated using the formula f_k = μ_k * m * g, where μ_k is the coefficient of kinetic friction. Plugging in the values given in the problem, we get f_k = 0.03 * 59.0 kg * 9.8 m/s² = 17.4 N. So if the applied force of 57.9 N is maintained, then the acceleration of the block would be a = (57.9 N - 17.4 N) / 59.0 kg = 0.69 m/s².