During a tug-of-war, team A pulls on team B by applying a force of 1390 N to the rope between them. How much work does team A do
1 answer:
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The final velocity of the skater is 2.34 m/s forward
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.
Before the ball is thrown, the total momentum is:
where
M = 61 kg is the mass of the skater
m = 4.0 kg is the mass of the ball
u = 2.5 m/s (forward) is the combined velocity of the skater and the ball
After, the ball is thrown at twice the velocity, so the final total momentum is
where
V is the final velocity of the skater
v = 2(2.5) = 5.0 is the final velocity of the ball
Since the total momentum must be conserved, we can write
So, the skater is moving at 2.34 m/s (forward) after the shot.
Learn more about momentum:
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To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have
Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s