1The normal force acting on a book of mass 1.40 kg sitting on a shelf sloped 11.

Question

1The normal force acting on a book of mass 1.40 kg sitting on a shelf sloped 11.0o from the horizontal is

a) 13,7 N

b) 13,5N

c) 14,2N

d) 14,3N

e) 14,0N

A 200 N sled is pulled up a 30° (from horizontal) slope at a constant speed by a force of 150 N. What is the coefficient of kinetic friction between sled and slope?

a) 0.53

b) 0.15

c) 0.29

d) 0.87

Tension adjusts so

a) the net force is zero

b)the vertical component always equals mg.

c)the force is always one-half the weight

d)he force is constant along the wire.

Explanation / Answer

Ques 2.

Because the sled is moving at a constant speed, there is no net acceleration. No net acceleration means that the sum of the forces acting on the sled equal zero. There are three forces acting on the sled: The 150N force directed up the slope, the gravitational force (m x g) directed down the slope, and friction (FF) which in this case is directed down the slope. On a horizontal surface the gravitational force would be equal to (m x g). On the incline only a component of the gravitational force is acting on the sled (m x g x sin30). The frictional force on a horizontal surface is equal to (Uk x n) where (n) is the normal force, and is equal to (m x g). On the incline the normal force is equal to (m x g x cos30). So the frictional force on the incline is equal to
(Uk x m x g x cos30) Now that we have defined all the forces acting on the sled, we can set them equal to zero and solve for our unknown.

150N - (m x g x sin30) - (Uk x m x g x cos30) = 0

Solve for Uk

Uk = [ 150N - (m x g x sin30)] / (m x g x cos30)

(m x g = 200N)

Uk = [ 150N - (200 x sin30)] / (200 N x cos30) = 0.288

The net force is zero.

Please post the other question separately.

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