Answer:
The work done will be 57.15 J
Explanation:
Given that,
Mass = 12 kg
Distance = 3 m
Force = 22 N
Angle = 30°
We need to calculate the work done
The work done is defined as,
Where, F = force
d = displacement
Put the value into the formula
Hence, The work done will be 57.15 J
Answer:
h = 120 m
h = 60 m
when rounded to two significant digits.
Explanation:
height of the higher ball drop point
h = ½gt² = ½(9.8)5.0² = 122.5 m
lower ball drop point
h = ½(9.8)(5.0 - 1.5)² = 60.025
θ = angle of the incline surface from the horizontal surface = 25⁰
μ = Coefficient of friction = 0.15
m = mass of the person = 65 kg
= kinetic frictional force acting on the person as he slides down
mg = weight of the person acting in down direction
= normal force by the incline surface on the person
the free body diagram showing the forces acting on the person is given as
Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
