Question

In this system, the acceleration of mass 1 is down the ramp. The masses are given an initial velocity of 2 m/s up the ramp. Eventually, the velocity of the masses is 4 m/s down the ramp. Mass 1 is 4 kg and mass 2 is 1 kg. There is friction between mass 1 and the surface it is on. The coefficient of kinetic friction is 0.20. The masses achieve their final velocity after mass 1 travels 6 meters down the ramp. The ramp angle is unknown. m. e ma

How much work would gravity do on mass 1?

A, 240cosθ

B, (240 J)sinθcosθ

C, 240sinθ

D, (240 J)

How much work would friction do on mass 1?

A, -48cosθ

B, -48sinθ

C, 48

D, 48sinθ

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Question 3 1 pts
How much work would tension do on mass 1?

Group of answer choices

(6)T

-(6)T

(6)Tsinθ

-(6)Tcosθ

How much work would gravity do on mass 2?

A, 60

B, 60cosθ

C, -60sinθ

D, -60

How much work would tension do on mass 2?

A, -(6)T

B, (6)Tsinθ

C, (6)T

D,(6)Tcosθ

What is the total work done on the system of both masses?

A, -(240)sinθ – (48)cosθ – 60

B, -(240)sinθ – (48)cosθ – 60 + (12)T

C, (240)sinθ – (48)cosθ – 60 – (12)T

D, (240)sinθ – (48)cosθ – 60

What is the change in the kinetic energy of the system of masses?

What is the angle of the ramp? (in degrees)

Answer 1

m1 = 4 kg, m2 = 1 kg, g = 10 m/s2; $\mu$ = 0.2; d = 6m; initial speeds u = 2 m/s ; final speeds v = 4 m/s

and let the tension be T. When the force and displacement are in same direction then work is positive. If force and displacement are in opposite direction then work done is negative.

d = 6 m is given down the incline but the friction will also do some work when m1 is going upwards. Probably its neglected as per the given options.

The component of gravitational force on mass m1 along the inclined plane is m1g sin$\theta$.

The work done by the gravitational force on mass m1 during its motion on inclined plane is Wg1 = m1g sin$\theta$ d

Wg1 = m1g sin$\theta$ d = 4 x 10 x sin$\theta$ x 6 = 240 sin$\theta$   

The work done by the frictional force on mass m1 during its motion on inclined plane is Wf1 = - $\mu$ m1g cos$\theta$ d

Wf1 = - $\mu$ m1g cos$\theta$ d = - 0.2 x 4 x 10 x cos$\theta$ x 6 = - 48 cos$\theta$

The work done by the tension in string on mass m1 during its motion on inclined plane is WT1 = - T d = - 6 T

The work done by the gravitational force on mass m2 during its motion is Wg2 = - m2g d = - 60

The work done by the tension in string on mass m2 during its motion is WT2 = T d = 6 T

The total work done on the system of masses is W = 240 sin$\theta$ - 48 cos$\theta$ - 6 T - 60 + 6 T

W = 240 sin$\theta$ - 48 cos$\theta$ - 60

The initial kinetic energy of the system is Ki = 1/2 (m1 + m2) u2 = 1/2 x 5 x 22 = 10J

The initial kinetic energy of the system is Kf = 1/2 (m1 + m2) v2 = 1/2 x 5 x 42 = 40J

The change in kinetic energy = Kf - Ki = 40 - 10 = 30 J

From work energy theorem, the total work done on the system of masses is change in kinetic energy

240 sin$\theta$ - 48 cos$\theta$ - 60 = 30 => 40 sin$\theta$ - 8 cos$\theta$ = 15 => 40 sin$\theta$ - 8 cos$\theta$ = 15

=> 40 sin$\theta$ - 15 = 8 cos$\theta$ => 40 sin$\theta$ - 15 = 8 [1 - (sin$\theta$)2]1/2 => (40 sin$\theta$ - 15)2 = 64 [1 - (sin$\theta$)2]

1600 sin2$\theta$ - 1200 sin$\theta$ + 225 = 64 - 64 sin2$\theta$ => 1664 sin2$\theta$ - 1200 sin$\theta$ + 161 = 0

sin$\theta$ = 0.543 or 0.178

$\theta$ = 32.89o or 10.25o