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3 years ago

14

A 200-kilogram wood crate is pushed across a wood floor How much force will be needed to move the box at a constant velocity of

0.5 m/s

1 answer:

lakkis [162]3 years ago

4 0

Answer:

100N

Explanation:

F = m x a

m = 200kg

a = 0.5m/s

F = 200 x 0.5

F = 100

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A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s

Alisiya [41]

Answer:

$v_{ic}=92.53 m/s$

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

<u>Conservation of momentum</u>

$p_{i}=p_{f}$

$m_{c}v_{ic}=(m_{c}+m_{w})V$ (1)

m(c) is the mass of stick clay
m(w) is the mass of the wooden block
v(ic) is the initial velocity of clay
V is the final velocity of the system clay plus wood.

<u>Conservation of total energy</u>

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

$\Delta E=W$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd$

We can find V from this equation:

$V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s$

Now, let's put V into the equation (1) and find v(ic)

$v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s$

I hope it helps you!

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