Question

A block with a mass m = 5 kg slides down an inclined plane with an angle θ = 37°. The block maintains a constant acceleration a = 5.6 m/s2 . (sin37° = 0.6, cos37° = 0.8). The coefficient of kinetic friction between the block and the inclined surface is 0.05. What is the normal force on the block?

Answer 1

Answer:1148.6 N

Step-by-step explanation:

Given

mass of block =5 kg

Inclination angle $\theta =37^{\circ}$

constant acceleration(a)$=5.6m/s^2$

$\mu$ be the coefficient of kinetic friction =0.05

we know friction force$=\mu N$

forces acting on block is $mgsin\theta$down the inclined  plane

friction force acting parallel to surface opposing $mgsin\theta$

Therefore

$mgsin\theta - \mu N=ma$

$\frac{m\left ( gsin\theta +a\right )}{\mu }=N$

$N=\frac{5\left ( 9.81\times 0.6+5.6\right )}{0.05}$

N=1148.6 N