A machine pulls a 46 kg trunk 3.0 m up a 42o ramp at constant velocity, with the machine's force on the trunk directed parallel
2 answers:
Answer:
the answers are
W = 1271.256 J
= 361.81 J
Explanation:
The mass of the trunk (m) = 46kg
angle between the ramp and the horizontal (θ) = 42°
The trunk is pulled at a displacement (d) up the ramp = 3m
The coefficient of kinetic friction between the trunk and the ramp = 0.36
The trunk is moving with a constant velocity therefore the net force on it is zero, therefore the force required to move the trunk must be equal to the summation of forces opposing the trunk
The two forces opposing the trunk are
- the gravitational force directed down the ramp
and
- the frictional force between the ramp and the trunk
We have to calculate the machine's force which is equal to sum of the
=
θ
μ× N
N = mgcosθ
μ
θ
F = mg (sinθ + μcosθ)
F = 46× 9.8 (sin42 + 0.36×cos42)
F= 450.8 (0.67 +0.27)
F = 450.8 × 0.94
F = 423.752N
to calculate the workdone on the trunk by the machine force
The workdone on the trunk is = W = F × dcosΘ
Θ = 0° because the trunk is directed parallel to the ramp
W =423.752× 3 cos 0
W = 423.752 ×3
W = 1271.256 J
(b) the increase in thermal energy of the trunk and the ramp
Friction converts mechanical energy into thermal energy, so multiplying the frictional force with the distance gives the thermal energy generated by the trunk
× d
= μθ ×d
= 0.36 ×46×9.8×cos42×3
= 361.81 J
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The value of normal force as the slider passes point B is
- 6 mg
The value of h when the normal force is zero
- 3R/2
The normal force is calculated using the work energy principle which is applied as below
K₁ + U₁ = K₂
k represents kinetic energy
U represents potential energy
the subscripts 1,2 , and 3 = a, b, and c
for 1 to 2
K₁ + W₁ = K₂
0 + mg(h + R) = 0.5mv²₂
g(h + R) = 0.5v²₂
v²₂ = 2g(1.5R + R)
v²₂ = 2g(2.5R)
v²₂ = 5gR
Using summation of forces at B
Normal force, N = ma + mg
N = m(a + g)
N = m(v²₂/R + g)
N = m(5gR/R + g)
N = 6mg
for 1 to 3
K₁ + W₁ = K₃ + W₃
0 + mgh = 0.5mv²₃ + mgR
gh = 0.5v²₃ + gR
0.5v²₃ = gh - gR
v²₃ = 2g(h - R)
at C
for normal force to be zero
ma = mg
v²₃/R = g
v²₃ = gR
and v²₃ = 2g(h - R)
gR = 2gh - 2gR
gR + 2gR = 2gh
3gR = 2gh
3R/2 = h
Learn more about normal force at:
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