Problem 1:
A block of mass 6 kg is released from point A and it slides down the incline ( = 30°) where the
coefficient of kinetic friction μ is 0.18. It goes 5.0 m and hits a spring with a spring constant k=700 N/m.
During the compression and the reaction of the spring, we assume that the bloc is moving on a frictionless
surface (μ = 0).
a. How far is the spring compressed?
b. What would be the velocity of the bloc when it is exactly at 1.5 m on its way up after the
rebound?
Answers
Explanation:
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Given :
m = 6kg
θ = 30°
μ = 0.18
l = 5m
k = 700N/m
To find :
a) How far is the spring compressed?
b) What would be the velocity of the bloc when it is exactly at 1.5 m on its way up after the rebound?
Solution :
a) Decrease in gravitational potential energy of block
PE = mgh
PE = 6 × 9.8 × 5sin30°
PE = 147J
Force acting on clock in perpendicular direction to surface are normal force N and component of gravitational force (mg) perpendicular to the inclined surface (mgcosθ)
N = mgcosθ
N = 6 × 9.8 × cos30°
N = 50.92
Frictional force acting on block
F = μN
F = 0.18 × 50.92
F = 9.1656N
Work done against frictional force
W = F.s = Fscosθ
W = 9.1656 × 5 cos0°
W = 45.828J
KE = PE - W
KE = 147 - 45.828
KE = 101.18J
On hitting the spring, block compress the spring and block's kinetic energy is stored into potential energy of spring.
PE = 1/2 × kx²
101.18 = 1/2 × 700 x²
x = 0.5377m = 0.54m
Spring is compressed by x = 0.54m = 53.77cm
b) Increase in gravitational energy of the block at point C
PE = mgh
PE = 6 × 9.8 × 1.5sin30°
PE = 44.1J
Work done against frictional force
W = F.s = Fscosθ
W = 9.1656 × 1.5 × cos0°
W = 13.75J
Kinetic energy left with block
KE = KEi - PE - W
KE = 101.18 - 44.1 - 13.75
KE = 43.33J
Velocity of block at point C
KE = 1/2 ×mv²
43.33 = 1/2 × 6 × v²
v = 3.8m/s
Velocity of block when it is 1.5m on its way up the inclined plane v = 3.8m/s