A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
0 N and is directed at 37° above the horizontal. What is the work done by this force?
2 answers:
Given Information:
Horizontal distance = d = 6.0 m
Force = F = 20 N
Angle = θ = 37°
Required Information:
Work done = W = ?
Answer:
Work done = 95.84 Joules
Explanation:
The work done on toy cart is given by
W = Fdcos(θ)
Where F is the force applied, d is the distance by which the toy cart has been displaced, and θ is the angle between F and horizontal floor.
As you can see in the attached diagram, The force pulling the cart makes an angle of 37 with respect to the horizontal surface.
W = 20*6*cos(37°)
W = 95.84 Joules
Therefore, 95.84 Joules of work has been by a force of 20 N to displace the toy cart by 6 meters.
You might be interested in
Answer:
A. The work done during the process is W = 0
B. The value of heat transfer during the process Q = 442.83
Explanation:
Given Data
Initial pressure = 100 k pa
Initial temperature = 25 degree Celsius = 298 Kelvin
Final pressure = 300 k pa
Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.
⇒ =
------------- (1)
Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.
⇒ P ∝ T
⇒ =
⇒ Put all the values in the above formula we get the final temperature
⇒ =
× 298
⇒ = 894 Kelvin
(A). Work done during the process is given by W = P × ()
From equation (1), =
so work done W = P × 0 = 0
⇒ W = 0
Therefore the work done during the process is zero.
Heat transfer during the process is given by the formula Q = m (
-
)
Where m = mass of the gas = 1 kg
= specific heat at constant volume of nitrogen = 0.743
Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )
⇒ Q = 442.83
Therefore the value of heat transfer during the process Q = 442.83