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Marrrta [24]
3 years ago
7

A friend on skis stands still on level ground, with dry 0.00°C with a coefficient of static friction of 0.0300. How hard would y

ou have to push your friend in the forward horizontal direction to move him if your friend weighs 685 N?​
Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

20.6 N

Explanation:

Friction equals normal force times coefficient of friction.

F = Fn μ

On level ground, normal force equal weight.

Fn = W

Therefore:

F = W μ

F = (685 N) (0.0300)

F = 20.6 N

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A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie
Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

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3 years ago
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