A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti
1 answer:
Answer:
<u></u>
Explanation:
Normal reaction from 40 kg slab on 10 kg block
M × g = 10 × 9.8 = 98 N
Static frictional force = 98 × 0.7 N
Static frictional force = 68.6 N is less than 100 N applied
10 kg block will slide on 40 kg slab and net force on it
= 100 N - kinetic friction
= 100 - 39.2
= 60.8 N
Frictional force on 40 kg slab by 10 kg block = 98 × 0.4
Frictional force on 40 kg slab by 10 kg block = 39.2 N
40 kg slab will move with =
You might be interested in
at the lowest point in the trajector, the kinetic energy of the bob is 300 J.
Explanation:
The total mechanical energy of the bob at any point of its motion is given by
Where
is the kinetic energy, where
m is the mass of the bob
v is its speed
is the gravitational potential energy, where
g is the acceleration of gravity
h is the height of the bob, measured with respect to the lowest point of the trajector
In absence of friction, the total mechanical energy E remains constant. So we have:
- When the bob swings upward, the PE increases (because h increases) and the KE decreases (so the speed decreases). At the highest point in the trajector, the speed of the bob is zero (v=0), so its KE is also zero and all the mechanical energy is potential energy: U = 300 J
- When the bob swings downward, the PE decreases (because h decreases) and the KE increases (so the speed increases). At the lowest point in the trajectory, the height has become zero (h=0), so the PE is zero and all the mechanical energy is kinetic energy: KE = 300 J
Therefore, at the lowest point in the trajector, the kinetic energy of the bob is 300 J.
Learn more about kinetic energy:
#LearnwithBrainly
Answer:
The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string
ΔP = F × Δt
Explanation:
The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body
ΔP = F × Δt
The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'
P = m × v
Tension is the axial pulling force of a string
T = Axial Force, F
The tension used in Tennis Racket strings is between 40 to 65 lbs.
When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized
When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.
However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.