HELP! block is released from rest and slides down a frictionless ramp inclined at 30' from the horizontal. When the block reache
s the bottom the block-Earth system has mechanical energy Eo The
experiment is repeated, but now horizontal and vertical forces of magnitude F are exerted on the block while it slides, as shown above. When the block reaches the bottom, the mechanical energy
of the block-Earth system
A) is greater than Eo
B)is equal to Eo
C)is less than Eo
D)cannot be determined without knowing
experiment is repeated, but now horizontal and vertical forces of magnitude F are exerted on the block while it slides, as shown above. When the block reaches the bottom, the mechanical energy
of the block-Earth system
A) is greater than Eo
B)is equal to Eo
C)is less than Eo
D)cannot be determined without knowing
1 answer:
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Answer:
The force of car 3 on car 2 ≈ 1810.82 N
Explanation:
The equation for the change in momentum of the two cars are;
Conservation of linear momentum
150( 2.2 - v) = 265(1.5-v)
150 × 2.2 - 265×1.5 = (150+265)v
150 × 2.2 - 265×1.5 = -67.5 = 415×v
∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East
The impulse of the net force is the amount of momentum change experienced given by the equation;
Impulse force = -
Where;
= The final velocity
= The initial velocity
For the the 265 kg mass, we have;
= 0.1627 m/s
= 1.5 m/s
Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s
The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J
Whereby the distance moved in one second is 0.1627 m, we have;
Work done = Force × Distance = Force × 0.1627 = 294.62
Force = 294.62/0.1627 = 1810.82 N.