A 50-kg box is being pushed a distance of 8.0 m across the floor by a force P with arrow whose magnitude is 159 N. The force P w
1 answer:
Answer:
Wp = 1,272 J
Wf = -940.8 J
Wn = 0
Wg = 0
Explanation:
- Applying the definition of work, as the product of the component of the force applied, along the direction of the displacement, times the displacement, we find that due to the normal force is always perpendicular to the surface, it does no work, as it has not a component in the direction of the displacement, so Wn = 0.
- As the weight goes directly downward, it has no component in the direction of the displacement either, so Wg = 0.
- We can get the work done by the force applied P, simply as follows:
- Finally, we have the work done by the force of friction that always opposes to the displacement, so it has negative sign.
- The frictional force , can be written as follows:
where μk = coefficient of kinetic friction = 0.24
Fn = Normal Force
- In this case, since the surface is level and horizontal, and there is no acceleration of the box in the vertical direction, this means that the normal force (in magnitude) must be equal to the weight:
- Fn = m*g = 50 kg * 9.8 m/s2 = 490 N
- Replacing these values in (2), we get:
- Applying the definition of work, we can get the work done by the frictional force, as follows:
You might be interested in
Answer
given,
distance of first satellite = 48,000 Km
distance of second satellite = 64,000 Km
orbital period = 6.39 day
Using equation of time period
now, from the above equation we can say that only variable is Time period and r is the radii of orbit.
from the first satellite
for second satellite