As friction force is constant , we have
Wfr = Ffrl = (5.00N)(1.00m) = 5.00J
DU = 0 - mgy = 0 - (3.00kg)(9.80m/s2)(1.00m)sin30.0° = -14.70J
DK = 1/2mvf2 - 0
Combining the three equations, we get
vf2 = 6.47(m/s)2, which gives vf = 2.54m/s.
QUIZ 4
1.A 3.00-kg crate slides down a ramp at a loading dock. the ramp is 1.00m in length and inclined at an angle of 30.0°, as shown in Figure. The crate starts from rest at the top, experiences a constant frictional force of magnitude 5.00N, and continues to move a short distance on the flat floor. Use energy methods to determine the speed of the crate when it reaches the bottom of the ramp.
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Solution. We have DE = DK + DU =
-Wfr
As friction force is constant , we have Wfr = Ffrl = (5.00N)(1.00m) = 5.00J DU = 0 - mgy = 0 - (3.00kg)(9.80m/s2)(1.00m)sin30.0° = -14.70J DK = 1/2mvf2 - 0 Combining the three equations, we get vf2 = 6.47(m/s)2, which gives vf = 2.54m/s.
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2. A block of mass m1 = 1.60kg initially moving to right with a speed of 4.00m/s on a frictionless horizontal track collides with a second block of mass m2 = 2.10kg moving to the left with a speed of 2.50m/s. After the collision m is moving to the left with a speed of 3.00m/s, determine the speed of m2.
Solution. Using conservation of momentum, we have
m1v1 + m2v2 = m1v1' +m2v2'
(1.60kg)(4.00m/s) + (2.10kg)(-2.50m/s) = (1.60kg)((-3.00m/s) + (2.10kg)v2'
which gives v2' = 2.83m/s to the right.
Note that we take the velocity to the right as positive, and the one to the left as negative.