Question

A block is pulled along a rough horizontal surface by force of magnitude 100 Newtons, at an angle of 30 degrees as shown. The block has a mass M=20.0 kg, and the coefficient of kinetic friction between the block and the surface is 0.50.

a)The horizontal component of the tension, T, is

b)The normal force exerted upward by the floor on the block has a magnitude of

c)If the block moves 10.0 m to the right, the work done by the force of gravity on the block is

d) Assuming the block starts from rest and is dragged 10.0 meters to the right, the block’s kinetic energy will be:

Answer 1

the horizontal force acting on the block is

Fx = ma

F cos thea - uk N = ma

T- ukN = F

T = F + uk N = 100 N + ( 0.5) ( 20) ( 9.8)

= 198 N

(b)

the net vertical force acting on the block is zero

Fy = 0

N+ F sin theta - mg = 0

N = mg- F sin theta

(c)

the work done by force of gravity

W = F cos thea s

= 100 cos 30 *10 m

= 866 m

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the finlal speed of the block is

1/2 mvf^2 = 1/2 mvi^2 - fk d + WF

vf = root vi^2 + 2/m ( - uk N d + F d)

= root ( 0 + 2/20 kg ( - 0.5 ( 20 kg ) ( 9.8) ( 10) + 100 ( 10 m)

= 1.414 m/s

the kinetic energy of the block is

KE = 1/2 m vf^2 = 1/2 * 20 kg ( 1.414 m/s)^2 = 20 J