To be referenced, it would be true
Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J
Answer:
Action: Gravity pulls on the ball.
Reaction: The ball falls to the ground.
Explanation:
Answer:3.31m/s², to the right
Explanation:
According to the law of conservation of momentum of a body, change in momentum of bodies before collision is equal to the change in momentum after collision.
Momentum = mass × velocity
M1 and M2 be the masses of the first and second skaters respectively
Let u1 and u2 be the velocities of the first and second skaters respectively.
v be their common velocity after collision
M1 = 77kg M2 = 66kg u1 = 4m/s² u2 = 2.5m/s²
According to the law we have
M1u1 + M2u2 = (M1+M2)v
77(4) + 66(2.5) = (77+66)v
308 + 165 = 143v
V = 473/143
V = 3.31m/s²
Their velocity after collision will become 3.31m/s²
They will both move towards the right after collision because the mass of the body moving to the right is higher than the other mass and the mass is also moving at a higher velocity than the other.
