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3 years ago

14

A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at

time t=0, a horizontal force Fx(t)=β−αt is applied to the block, where α = 6.00 N/s and β = 4.00 N.
What is the largest positive value of x reached by the block?
How long does it take the block to reach this point, starting from t = 0?
What is the magnitude of the force when the block is at this value of x?

1 answer:

Allisa [31]3 years ago

8 0

Answer:

x = 1,185 m , t = 4/3 s , F = - 4 N

Explanation:

For this exercise we use Newton's second law

F = m a = m dv /dt

β - α t = m dv / dt

dv = (β – α t) dt

We integrate

v = β t - ½ α t²

We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

v-v₀ = β t - ½ α t²

the farthest point of the body is when v = v₀ = 0

0 = β t - ½ α t²

t = 2 β / α

t = 2 4/6

t = 4/3 s

Let's find the distance at this time

v = dx / dt

dx / dt = v₀ + β t - ½ α t2

dx = (v₀ + β t - ½ α t2) dt

We integrate

x = v₀ t + ½ β t - ½ 1/3 α t³

x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³

The body comes out of rest

x = 3.5556 - 2.37

x = 1,185 m

The value of force is

F = β - α t

F = 4 - 6 4/3

F = - 4 N

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