Answer:
The work done on the system can be accounted for by;
Both
and 
Explanation:
The speed of the crate = Constant
Therefore, the acceleration of the crate = 0 m/s²
The net force applied to the crate,
= 0
Therefore, the force of with which the crate is pulled = The force resisting the upward motion of the crate
However, we have;
The force resisting the upward motion of the crate = The weight of the crate + The frictional resistance of the ramp due to the surface contact between the ramp and the crate
The work done on the system = The energy to balance the resisting force = The weight of the crate × The height the crate is raised + The heat generated as internal energy to the system
The weight of the crate × The height the crate is raised = Gravitational Potential Energy = 
The heat generated as internal energy to the system = 
Therefore;
The work done on the system =
+
.
F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
There is a possibility but not extremely likely
A = delta v over delta t delta v is calculated with final velocity less initial velocity so delta v is 14 - 6 that is 8m/s and delta t is calculated with final time less initial time as initial always is 0 then is 1 - 0 that is 1 then a = 8m/s over 1 that is 8 then the acceleration is 8m/s^2 (remember that is squared.)
Answer:
True
Explanation:
Examples of numerical data are height, weight, age