Question

In: Physics

Is work done when dragging a mass up a hill the same as the change in...

Is work done when dragging a mass up a hill the same as the change in potential energy?

Is work the same as energy? I am confused as to why

Solutions

Expert Solution

Work is not the same as energy.

Work done a a body is equal to the net change in the kinetic energy of the body.

In your case if the body is dragged uphill by a constant velocity, only then will the work done be same as change in potential energy.

Take for example that friction is not acting on body

Initial velocity of block = v m/s

Initial height= 0 m

final velocity= v/ms

final height = h m

Work done is lifting body by height h metre= F d = (-mg)*h = -mgh (-mg because gravity acts downwards)

Change in Kinetic energy = (1/2) m (v12 - v22 ) = (1/2)m (v2-v2)=0

Change in Kinetic energy= change in Potential energy + Work done

=> Work done = - Change in potential energy

=> Work done= -mg (h2-h1)= -mgh

If the velocity of body doesnt remain constant , then

Work done= Change in kinetic energy- Change in potential energy- Work done by frictional forces.

Thus work and energy are not the same thing but change in energy of system gives work done on it.


Related Solutions

How high a hill can a car coast up (engine disengaged) if work done by friction...
How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 104 km/h? 42.6 m If, in actuality, a 870–kg car with an initial speed of 104 km/h is observed to coast up a hill to a height 31.9 m above its starting point, how much thermal energy was generated by friction? 9.11×104 J What is the average force of friction if the hill has a slope 7.7°...
You push a box of mass 19.1 kg with your car up to an icy hill...
You push a box of mass 19.1 kg with your car up to an icy hill slope of irregular shape to a height 5.9 m. The box has a speed 12.9 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) to a flat area before sliding into a box larger box of mass 13 kg. The boxes then fall off a sheer cliff together to the...
You push a box of mass 17.4 kg with your car up to an icy hill...
You push a box of mass 17.4 kg with your car up to an icy hill slope of irregular shape to a height 5.1 m. The box has a speed 11.8 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) to a flat area before elastically colliding with a smaller box of mass 12 kg. The boxes then fall off a sheer cliff individually to the...
Calculate the work done by the adiabatic expansion between the same volumes used in the isothermal...
Calculate the work done by the adiabatic expansion between the same volumes used in the isothermal expansion: 2 m3 to 5 m3 for both the a.) irreversible and b.) reversible processes. Use a monoatomic ideal gas: CV=3R/2 (bar above CV); P1 = 5 Pa; take T1 to be 300K
Consider rider on a typical bicycle pedalling up a hill. Specifically, the case when the cyclist...
Consider rider on a typical bicycle pedalling up a hill. Specifically, the case when the cyclist is using their body weight alone to apply the pedalling force. Draw a diagram of the situation that you can analyse the system and answer the following questions: a) Does the mass of the rider affect the acceleration of the bike (and rider) up the hill? b) Is there an optimum ratio from pedal crank radius to rear wheel (including the chain drive ratio)...
Work done.
A ring of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
Work done by force.
A force acts on a 2 Kg object so that its position is given as a function of times as x=3t²+5. What is the work done by this force in the first 5 seconds?
How can I decide in a thermodinamic process from pressure and/or volume change if work done...
How can I decide in a thermodinamic process from pressure and/or volume change if work done by the system or work done on the system? Since I know if work done on the system dU=dW+dQ, work done by the system dU=dQ-dW. However for me very difficult to decide whether in the final answer work will be neative or positive. Is it true that if it is compressionw:work done by the system (work negative) and expansion:work done on the systen (work...
calculate the work done when: a) a 55 N force acting at an angle of 28...
calculate the work done when: a) a 55 N force acting at an angle of 28 degrees to the horizontal pushes an object 5m horizontally. b) a spring of spring constant 45n/m is stretched by 32cm.
Carefully work through the sample calculations for this lab ON PAPER. Then, set up the same...
Carefully work through the sample calculations for this lab ON PAPER. Then, set up the same equations on a separate sheet of paper, but replace the values in the ‘Sample Calculations’ folder with the ones provided below and complete the calculations: Mass of Zinc = 0.0928 g H2 volume reading from the eudiometer tube = 35.98 mL Water bath temperature = 19.9 °C Barometric pressure = 768.2 mm Hg Level (pressure) difference from the meter stick = 41.20 cm H2O...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT