Question

A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. What isthe acceleration of the skier?(Unit = m/s?)Enter

Answer 1

Free body diagram:

Apply second law of newton:

∑F = m*a

-Wx+ F1- FF = m*a

N-wy = 0

w= m g

m=72.5 kg

g= 10N

w= 72.5 * 10 = 725 N

wx = w sin 21.7 = 725 sin 21.7 = 268 N

wy= w cos 21.7= 725 cos 21.7 = 674 N

N= wy

N= 674 N

F1= 383N

FF= u* N = 0.120 * 674N = 81 N

-Wx+ F1- FF = m*a

-268+383N-81 = 72.5 * a

34N = 72.5 kg a

34 kgm/s2 / 72.5 kg = a

0.47 m/s2 = a