Question

A 50.0 kg crate is being pulled along a horizontal, smooth surface. The pulling force is 10.0 N and is directed 20.0 degree above the horizontal. Find the normal force the crate and the magnitude of the acceleration of the crate. A. 250 N, 0.0684 m/s^2B. 487 N, 0.188 m/s^2 C. 525 N, 0.200 m/s^2 D. 825 N, 0.376 m/s^2 E. 494 N, 0.0728 m/s^2

Answer 1

Explanation:

It is given that,

Mass of the crate, m = 50 kg

Force acting on the crate, F = 10 N

Angle with horizontal, $\theta=20^{\circ}$

Let N is the normal force acting on the crate. Using the free body diagram of the crate. It is clear that,

$N=mg-F\ sin\theta$

$N=50\times 9.8-10\ sin(20)$

N = 486.57 N

or

N = 487 N

If a is the acceleration of the crate. The horizontal component of force is balanced by the applied forces as :

$ma=F\ cos\theta$

$a=\dfrac{F\ cos\theta}{m}$

$a=\dfrac{10\times \ cos(20)}{50}$

$a=0.1879\ m/s^2$

or

$a=0.188\ m/s^2$

So, the normal force the crate and the magnitude of the acceleration of the crate is 487 N and $0.188\ m/s^2$ respectively.