`A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s2 the resulting acceleration of the slab will be A.0.98 m/s2B.1.47 m/s2 C.1.52 m/s2D.6.1 m/s2
Question
`A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s2 the resulting acceleration of the slab will be A.0.98 m/s2B.1.47 m/s2 C.1.52 m/s2D.6.1 m/s2
Solution
First, we need to determine the maximum static friction force that can act on the 10 kg block. This is given by the equation:
F_max = μ_s * m * g
where μ_s is the static coefficient of friction, m is the mass of the block, and g is the acceleration due to gravity. Substituting the given values, we get:
F_max = 0.60 * 10 kg * 9.8 m/s² = 58.8 N
Since the applied force (100 N) is greater than the maximum static friction force (58.8 N), the block will start to slide on the slab. The friction force acting on the block when it is sliding is given by the kinetic friction force, which is:
F_k = μ_k * m * g
where μ_k is the kinetic coefficient of friction. Substituting the given values, we get:
F_k = 0.40 * 10 kg * 9.8 m/s² = 39.2 N
The net force acting on the slab is the difference between the applied force and the kinetic friction force, which is:
F_net = F_applied - F_k = 100 N - 39.2 N = 60.8 N
The acceleration of the slab is given by Newton's second law, F = ma, where F is the net force and m is the mass of the slab. Solving for a, we get:
a = F_net / m = 60.8 N / 40 kg = 1.52 m/s²
So, the resulting acceleration of the slab is 1.52 m/s², which corresponds to option C.
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