YouA block of mass 2.0kg is placed on a trolley of mass 5.0kg, moving horizontally. A force of 8.0Nis applied to the block which slides on the surface of the trolley. The frictional force between thetrolley and the ground is zero.The trolley accelerates at a rate of 1.0ms 2. What is the coefficient of dynamic friction between theblock and the trolley?A. 0.05B.0.15C. 0.25D. 0.35
Question
YouA block of mass 2.0kg is placed on a trolley of mass 5.0kg, moving horizontally. A force of 8.0Nis applied to the block which slides on the surface of the trolley. The frictional force between thetrolley and the ground is zero.The trolley accelerates at a rate of 1.0ms 2. What is the coefficient of dynamic friction between theblock and the trolley?A. 0.05B.0.15C. 0.25D. 0.35
Solution
To find the coefficient of dynamic friction, we first need to find the frictional force acting on the block.
We know that the net force acting on the system (block + trolley) is equal to the total mass of the system multiplied by the acceleration of the system.
So, F_net = (m_block + m_trolley) * a => F_net = (2.0kg + 5.0kg) * 1.0ms^-2 => F_net = 7.0N
The force applied on the block is 8.0N. Since the net force acting on the system is 7.0N, the frictional force (F_friction) acting on the block is the difference between the force applied on the block and the net force acting on the system.
So, F_friction = F_applied - F_net => F_friction = 8.0N - 7.0N => F_friction = 1.0N
The coefficient of dynamic friction (μ) is the ratio of the frictional force to the normal force. Since the block is not moving vertically, the normal force is equal to the weight of the block, which is m_block * g, where g is the acceleration due to gravity (approximately 9.8ms^-2).
So, μ = F_friction / (m_block * g) => μ = 1.0N / (2.0kg * 9.8ms^-2) => μ = 0.05
Therefore, the coefficient of dynamic friction between the block and the trolley is 0.05 (Option A).
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