Question

A 40 kg box lies on a plane as shown below. If the coefficient of friction (uk) between the box and the incline is 0.2, what is the acceleration of the box? 37°

Answer 1

ч W. Coso W. sing w

By force balance in the direction of inclined plane,

W*sin$\theta$ + Fk = m*a eq(1)

here, W = weight = m*g

m = mass of car = 40 kg

g = 9.81 /s^2

$\theta$ = angle of inclined plane = 37 deg

a = acceleration of car = ??

Fr = friction force applied on car = -$\mu$*N

N = normal reaction force = ??

given, $\mu$ = Co-efficient of kinetic friction = 0.20

from force balance in the perpendicular direction of inclined plane,

N - W*cos$\theta$ = 0 (since there is no acceleration in the perpendicular direction of inclined plane.)

N = m*g*cos$\theta$

So, from eq(1),

m*g*sin$\theta$ - $\mu$ *m*g*cos$\theta$ = m*a

a = g*sin$\theta$ - $\mu$ *g*cos$\theta$

a= 9.81*sin(37 deg) - 0.20*9.81*cos(37 deg)

a = 4.34 m/s^2

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