A string connected to a 10.0 kg mass resting on a table passes over a frictionle

Question

A string connected to a 10.0 kg mass resting on a table passes over a frictionless pulley and is connected to a 2.00 kg mass which rests on a surface inclined at an angle of 30.0 degree to the horizontal, (see the diagram below) Calculate the tension in the string when the 10.0 kg mass is hold stationary. Calculate the acceleration of the masses when they are moving, (magnitude only) Calculate the tension in the string while the masses are moving. A crate with a weight of 194 newtons rests on a level floor. A horizontal force of 8. 00 N is required to start the crate moving. A horizontal force of 2.00 N is required to keep the crate moving at a constant velocity. Calculate the coefficient of static friction. Calculate the coefficient of kinetic friction. A 2 00 kg metal block was placed on a table top. The metal block's coefficient of kinetic friction was 0.175. A horizontal force of 5.00 N was to the metal block. Calculate the magnitude of net force acting on the metal block. Calculate the magnitude of the metal block's acceleration. How long did it take to increase the block's speed from 2.50 m/s to 3.75 m/s? d) How far did the metal block travel while its speed increased from 2.50 m/s to 3.75 m/s? A 50 0 kg crate rests on a horizontal surface. The crate's coefficient of kinetic friction is 0.25. Two ropes are attached to the crate. A force of 200. newtons b 30 0 degree north of east is applied to one of the ropes and a force of 150. newtons b 20.0 degree north of east is applied to the other rope. Calculate the kinetic friction. Calculate the net force acting on the crate. Calculate the crate's acceleration (magnitude and direction). How long it take the cart to increase its speed from 6.24 m/s to 10.4 m/s? How far will the cart travel between times 1 = 3.00 seconds and t = 5.00 seconds if it started at time t = 0? A 0.500 kg puck sliding at 2.50 nvs on a frictionless surface moved on to a surface whose coefficient of kinetic friction was 0.0900. Calculate the magnitude of the pucks acceleration on the second surface. How far did the puck slide on the second surface before it stopped? How long after it moved on to the second surface did the puck stop?

Explanation / Answer

a) Force of gravity and force of weight are equivalent, and will cancel out. So, the net force is going to be calculated in only the x direction. Total FORCE = Force_applied - friction Total FORCE = Force_applied - coefficient of kinetic friction*Normal force Total FORCE = Force_applied - u_k*F_n remember normal force = gravity Total FORCE = Force_applied - u_k*mg Total FORCE = 5 N. - .175(2kg.*9.81m/s^2) Total FORCE = 1.57 N b) F = ma a= F/m a = 1.57 N./2 kg a = .783 m/s^2 c) acelleration = difference in velocity/time t = a/v t = .783 m/s^2 / (3.75 m/s - 2.5 m/s) t = .627 seconds d) look up the four kinematic equations. I believe one of them that will apply to this situations is (v_final)^2 = (v_initial)^2 + 2ax isolate for x: x= (v_f^2 - v_i^2)/2a x= (3.75^2-2.5^2)/(2*9.81) x = 13.7 meters

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