A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide freely along the smooth vertical guides BC and
DE. Determine the velocity of the block and plate just after impact. The coefficient of restitution between the block and the plate is e = 0.75. Also, find the maximum compression of the spring due to impact.The spring has an unstretched length of 600 mm
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate
1 answer:
Answer:
Check the explanation
Explanation:
<u>Given data </u>
Mass of the block A is Ma = 10kg
Mass of the plate is Mp = 5kg
Coefficient of restitution between the block and that of the plate is e=0.75
Unstretched length of the spring is 0.6m
From the conservation of energy, we have four motion of block A from (1) to (2)
Then accordingly, we will have the step by step calculation in the attached image below .
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Answer:
A) About newtons
B) 76.518 newtons
C) 111.834 newtons
Explanation:
A) , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:
B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.
C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.
Hope this helps!