20 October, 02:02

# 1. It takes 26.5 N to pull a 72 kg sled across snow. What is the coefficient of friction? 2. A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction is 0.8, what is the weight of the box?3. A rightward force of 246 N is applied to a 35 N suitcase to accelerate it across the floor. The coefficient of friction between the crate and the floor is 0.75. What is the acceleration of the crate?4. If a 1500 N force is exerted on a 200 kg statue to move it across the floor. If the coefficient of friction is 0.37, what is the crate's acceleration?

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1. 20 October, 02:54
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1) It takes 26.5 N to pull a 72 kg sled across snow. What is the coefficient of friction?

The 26.5N. is all friction.

Normal force from snow = (mg) = (72 x 9.8) = 705.6N.

µ = (26.5/705.6) = 0.0376

2) A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction is 0.8, what is the weight of the box?

The friction is 565N.

(565 / 0.8) = 706.25N. weight.

3) A rightward force of 246 N is applied to a 35 N suitcase to accelerate it across the floor. The coefficient of friction between the crate and the floor is 0.75. What is the acceleration of the crate?

(35/g) = mass of 3.57kg. (g=9.8 m/s^2).

(35 x 0.75) = 26.25N. friction.

(246 - 26.25) = net 219.76N. accelerating force.

Aceleration = (f/m) = (219.76/3.57) = 61.56m/s^2.

4) If a 1500 N force is exerted on a 200 kg statue to move it across the floor. If the coefficient of friction is 0.37, what is the crate's acceleration?

200 x g = normal force of 1,960N.

Friction = (1,960 x 0.37) = 725.2N.

Net accelerating force = (1500 - 725.2) = 774.8N.

Acceleration = (f/m) = (774.8/200) = 3.874m/sec^2