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SpyIntel [72]
3 years ago
8

A 50kg boy jumps off the back of a 4 kg skateboard moving forward. Find the skateboard’s velocity immediately after the boy jump

s, assuming that the skateboard’s initial velocity is 5 m/s and the boy’s velocity when jumping off the back is 2.5 m/s in the forward direction.
Physics
1 answer:
AleksAgata [21]3 years ago
6 0

The mass of the boy and skateboard is there to throw you off. X and Y velocities are independent of each other, meaning that the only things you need are the velocity of the skateboard and the velocity of the boy. The total velocity of the boy is 7.5 m/s because the boy would still be moving with the 5 m/s that the skateboard has if the boy were to just jump straight up. Since the boy jumps forward with a velocity of 2.5 m/s, you would add them together.

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The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
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Answer:

Magnitude of the Frictional force = (mv₀²)/2x₁

Explanation:

For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.

According to Newton's first law of motion

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x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)

a = ?

v² = u² + 2ax

0 = (v₀)² + 2ax₁

2ax₁ = - v₀²

a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)

Magnitude of the Frictional force = ma = (mv₀²)/2x₁

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