Question

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?

Answer 1

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