Question

A sled, which has a mass of m = 125 kg., is sitting on an icy horizontal surface. A rope is attached to the front end of the sled such that the angle between the rope and the horizontal is ɑ= 28.0°. A force of 585 N is applied to the rope, and as a result, the sled accelerates to the right at a rate of 3.30 m/s^2.
Required:
a. Draw the free body diagram showing all the forces acting on the sled.
b. What is the magnitude of the frictional force acting on this sled?
c. What is the magnitude of the normal force acting on the sled?
d. What is the coefficient of sliding friction between the sled and the icy horizontal surface?
e. What will be the displacement of this sled at the end of 5.0 seconds?

Answer 1

Answer:

Explanation:

b ) Net force = mass x acceleration

= 125 x 3.3 = 412.5 N

Forward force ( horizontal component of F ) = 585 cos 28

= 516.5 N

Net force = forward force - friction

412.5 = 516.5 - friction

friction = 516.5 - 412.5 = 104 N .

c ) Nornal force = R

Total upward force = R + 585 sin28

= R + 274.64

For balancing

R + 274.64 = mg

R + 274.64 = 585 x 9.8 = 5733

R = 5458.36 N

d ) coefficient of sliding friction be μ

μ R = friction

μ x 5458.36  = 104

μ = .019

e )

Displacement in 5 sec

s = ut + 1/2 a t²

= 0 + .5 x 3.3 x 5²= 41.25 m .