The answer is 0.025J.
W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J
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.
Explanation:
Take north to be positive and south to be negative.
a = (v − v₀) / t
a = (-4.5 m/s − 4.5 m/s) / 8 s
a = -1.125 m/s²
The acceleration is 1.125 m/s² south.
Answer:
The distance travel by block before coming to rest is 0.122 m
Explanation:
Given:
Mass of block 
kg
Initial speed of block 
Final speed of block 
Coefficient of kinetic friction 
Ramp inclined at angle 
28.4°
Using conservation of energy,
Work done by frictional force is equal to change in energy,
Where 
m
Therefore, the distance travel by block before coming to rest is 0.122 m
