A rope accelerates a 4.00 kg box on a horizontal surface, as shown. The tension
ID: 1437868 • Letter: A
Question
A rope accelerates a 4.00 kg box on a horizontal surface, as shown. The tension on the rope has a constant magnitude of 20.0 N and makes an angle Theta with the horizontal. Staring from rest, during a 2.50 meter displacement the work done by the tension is 25.0 J, and the box reaches a speed of 2.00 m/s. There is friction between the box and floor. Calculate the angle Theta between the rope and horizontal. Calculate the work done by the friction force during the same displacement interval of 2.50 m. The graph shows the force acting on an object as a function of displacement in the x-direction. What is the work done during the displacement from x = 0.0 to 3.0 m?Explanation / Answer
1)
m = 4.0 Kg
Tension in the Rope, T = 20.0 N
Force in the positive direction, = T*cos()
Friction Force, F = uk*(m*g - T*sin())
Work done = Force * displacement
Work done by tension, T*cos()*2.5 = 25.0
20.0 * cos() * 2.5 = 25
= 60o
Angle btw rope & horizontal, = 60o
Initial Velociy, u = 0
Final Velocity, v = 2.0 m/s
Displacement,s = 2.5 m
v^2 = u^2 + 2*a*s
2.0^2 = 0 + 2*a*2.5
a = 0.8 m/s^2
T*cos() - Fr = m*a
20*cos(60) - Fr = 4*0.8
Fr = 10 - 3.2
Fr = 6.8 N
Work done by Friction Force, Wf = 6.8 * 2.5 J
Work done by Friction Force, Wf = 17 J
2)
Area under the Force- Displacement Graph is work done.
W = 1/2*30*1.0 + 20*1.0 + 10*1.0 + 1/2*10*1.0 + 1/2*1.0*10 J
W = 55 J
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