3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t
1 answer:
Answer:
Work done is 12.3 J
Explanation:
We have,
Mass of puck, m = 0.35 kg
Force of friction acting on the puck when it slides is 0.15 N
Distance travelled by the puck is 82 m.
It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by
The work done on the puck is 12.3 J.
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Answer:
final velocity = 0.08585m/s
Explanation:
We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.
The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after
total linear momentum before = linear momentum of first train car + linear momentum of second train car
We know that linear momentum = mv
where,
m = mass
v = velocity
thus,
total linear momentum before = m₁v₁ + m₂v₂
m₁ = mass of first train car = 135,000kg
v₁ = velocity of first train car = 0.305m/s
m₂ = mass of first second car = 100,000kg
v₂ = velocity of second train car = −0.210m/s
Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.
total linear momentum before = m₁v₁ + m₂v₂
= 135,000x0.305 + 100,000x(−0.210)
= 135,000x0.305 - 100,000x0.210
= 20,175 kgm/s
Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).
Therefore, total linear momentum after = mv
m = m₁ + m₂ = 135,000 + 100,000 = 235,000
total linear momentum before = total linear momentum after
235,000v = 20,175
v =
= 0.08585m/s