Is there anyone that is good with physics that can help me with my homework?

Question

#1)In serving, a tennis player accelerates a 56kg tennis ball horizontally from rest to a speed of 30m/s. Assuming that the acceleration is uniform when the racquet force is applied over a distance of 0.5m, what is the magnitude of the force exerted on the ball by the racquet?

#2.)The coefficients of static and kinetic friction between a 50kg box and a horizontal surface are 0.500 and 0.400,respectively.
a.) What is the acceleration of the object if a 250N horizontal force is applied to the box?
b.) What is the acceleration if the 250N force is applied at an angle of 45 degrees above the horizontal?

#3.)A lawnmower is being pushed on a level lawn with a constant force of 200N at an angle of 30degrees downward from the horizontal. How far does the student push the mower in doing 1.44*10^3 Joules of work?

#4.)What is the maximum height reached by a 0.300 kg ball thrown vertically upward with an initial speed of 10.0m/s?

Answer 1

#1 use conservation of energy:

The work done by the force of the raquet is equivalent to the kinetic energy of the ball.

F*d=1/2*m*v^2
you have all of the variables except F.

F=.5*.56*30*30/.5
=504 N


#2
It will take a force of:
50*9.8*.5
=245N
to overcome the static friction

a) The net force accelerating the box is:
250-50*9.8*.4
=54
which =m*a
so a=54/50
=1.08 m/s^2

b) If the force is pushing downward it is an insufficient force to overcome static friction or keep the box moving against the kinetic friction. The box is at rest
If the force is pushing upward at a 45 degree angle, then the force of static friction is
(50*9.8-sqrt(2)*250)*.5
=156N
and the hoizontal component of the force is
Sqrt(2)*250=176
which is strong enough to overcome the static friction.
So acceleration again using
F=m*a=176-(156*.4/.5)
a=51.2/50
=1.024 m/s^2


j

Answer 2

Hint for #4: the mass of the ball is irrelevant.