Question

A 2.3kg box, starting from rest, is pushed up a ramp by a 10 N force parallel to the ramp. The ramp is 2.0 m long and tilted at 17 degrees. The speed of the box at the top of the ramp is 0.80m/s.How much work does the force do on the system?What is the change in the kinetic energy of the system?What is the change in the gravitational potential energy of the system?What is the change in the thermal energy of the system?

Answer 1

Part a)

`W = (10 N)(2 m) = 20 J`

Part b)

`\Delta K = 0.736 J`

Part c)

`\Delta U = 13.2 J`

Part d)

`U_(thermal) = 5.66 J`

Step-by-step explanation:

Part a)

Work done by the applied force is given by the formula

`W = F.d`

here we know that

`F = 10 N`

`d = 2 m`

`W = (10 N)(2 m) = 20 J`

Part b)

As we know that the box was at rest initially and then it is moving with speed 0.80 m/s

so here we can say

`\Delta K = (1)/(2)mv_f^2 - (1)/(2)mv_i^2`

`\Delta K = (1)/(2)(2.3)(0.80)^2 - 0`

`\Delta K = 0.736 J`

Part c)

Change in gravitational potential energy is given as

`\Delta U = - W_g`

`\Delta U = -(-mg sin\theta d)`

`\Delta U = (2.3)(9.81)(sin17)(2)`

`\Delta U = 13.2 J`

Part d)

Now by energy conservation law we can say that

Work done by external agent = change in kinetic energy + change in potential energy + thermal energy lost

so we have

`20 = 13.6 + 0.736 + U_(thermal)`

`U_(thermal) = 5.66 J`

Answer 2

(a). The work done on the system by force is 20 J.

(b). The change in kinetic energy of the system is 0.736 J.

(c). The change in the gravitational potential energy of the system is 13.18 J

(d). The thermal energy of the system is 6.84 J.

Step-by-step explanation:

Given that,

Mass of box = 2.3 kg

Force = 10 N

Length = 2.0 m

Angle = 17°

Speed = 0.80 m/s

(a). We need to calculate the work done

Using formula of work done

`W=F* d`

Put the value into the formula

`W=10*2.0`

`W=20\ J`

The work done on the system by force is 20 J.

(b). We need to calculate the change in kinetic energy of the system

Using formula of change of kinetic energy

`\Delta K.E=K.E_(f)-K.E_(i)`

`\Delta K.E=(1)/(2)mv^2-0`

Put the value into the formula

`\Delta K.E=(1)/(2)*2.3*(0.80)^2`

`\Delta K.E=0.736\ J`

The change in kinetic energy of the system is 0.736 J.

(c). We need to calculate the change in the gravitational potential energy of the system

Using formula of gravitational potential energy

`P.E=mgh\sin\theta`

Where, h = change in height

Put the value into the formula

`P.E=2.3*9.8*2.0\sin17`

`P.E=13.18\ J`

The change in the gravitational potential energy of the system is 13.18 J.

(d). We need to calculate the thermal energy of the system

Using formula of thermal energy

Work done=Change in kinetic energy+change in potential energy+change in thermal energy

`\Delta U_(th)=W-\Delta K.E+-Delta P.E`

Put the value into the formula

`\Delta U_(th)=20-0.736-13.18`

`\Delta U_(th)=6.084\ J`

The thermal energy of the system is 6.84 J.

Hence, (a). The work done on the system by force is 20 J.

(b). The change in kinetic energy of the system is 0.736 J.

(c). The change in the gravitational potential energy of the system is 13.18 J

(d). The thermal energy of the system is 6.84 J.