Ablock is released from rest and slides down an incline. the coefficient of sliding friction is 0.38 and the angle of inclination is 60.0°. use energy considerations to find how fast the block is sliding after it has traveled a distance of 38.7 cm along the incline.

2.27m/s

Explanation:

The coefficient of sliding friction = 0.38, the incline angle = 60o

Coefficient of sliding friction = frictional force/ force of normal (fn)

The normal force = mgcos60 where m is the mass of the body and g is acceleration due to gravity

Frictional force = coefficient of friction*fn = 0.38*9.81*m*cos 60 = 1.8639m

Force tending to push the body down the incline plane = mgsin60 = 8.496m

Net force on the body = force pushing downward - Frictional force

Net force = 8.496m - 1.8639m = 6.632m

Kinetic energy of the body = 1/2 mv^2 = work done by the net force pushing the body down = net force * distance travelled

1/2mv^2 = 6.632m * 0.387in meters

Cancel mass on both side leaves 1/2v^2 = 6.632*0.387

V^2 = 2.566584*2 = 5.133

V = √5.133 = 2.27m/s

explanation:

the masses of the objects and the distance between them

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