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A big frozen tofu turkey in a box of total mass m=5kg is pushed up an incline (f

ID: 2129553 • Letter: A

Question

A big frozen tofu turkey in a box of total mass m=5kg is pushed up an incline (frictionless) a vertical distance of h=3m. . The block is traveling at the same speed (v1) at the bottom (1) of the incline as at the top (2). The speed at the top is v2=3m/s. At top the box slide a distance d1=15 cm on a frictionless surface until it encounters a friction patch of width d2=50 cm. Then the box slides again on a frictionless surface until it encounters a spring with spring constant k=500 N/m. The box compresses the spring a distance d3=10 cm before it comes momentarily to rest. Again all surfaces are frictionless except of the rough patch For all the questions in this problem use g=10m/s2

(a) How much work is done by the pushing force pushing the block up the incline?

(b) What is the total work done on the block by all forces on its route up the frictionless incline (1->2)? Explain

Please explain the second question with care.. that's the one I have trouble understanding..

I'll not give life saver if there is no explanation for the second part


Explanation / Answer

a) work =KE+PE

since no change in velocity from 1 to 2. KE = 0

work = PE = mgh=5x10x3=150 J

b) Work Energy theorem states that, the net change in Kinetic Energy of a particle (system) is equivalent to the sum of work done by all types of external forces be it conservative or non-conservative.

Net Work done is zero.

Explanation : There are two forces acting on the body. One force which moves the body up the inclined plane. Work done by this force W1=F.S=mgh

F= force = weight of the body, S= displacement = height change of the body

Second force is gravity.

Work done by gravity W2 = -mgh (minus because gravity is acting downward and the movement of the block is upward direction, i.e against the gravitational force.)

(Note: Yes, work done by a force can be negative if the displacement of the object is opposite to the direction of the force acting)

Now, net work done = W1+W2 = 0

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