Question

1. A sliding block of mass m-0.25 kg is subject to a force of magnitude 4 N that makes an angle of p 30° with the horizontal surface. If the coefficient of kinetic friction between block and surface is 0.5, what is the resulting acceleration of the block along the surface?

Answer 1

Here, for the normal force

N = mg - F * sin(30)

N = 0.25 * 9.8 - 4 * sin(30)

N = 0.45 N

now , for the acceleration

m *a = F * cos(30) - u * N

0.25 *a = 4 * cos(30) - 0.5 * 0.45

a = 13 m/s^2

the acceleration of the block is 13 m/s^2