A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho
1 answer:
You might be interested in
Answer:
(A). The work done on the cart by the friction is -700 J.
(B). The work done on the cart by the gravitational force is 0 J.
(C). The work done on the cart by the shopper is 700 J.
(D). The force the shopper exerts 38.61 N.
(E). The total work done on the cart is zero.
Explanation:
Given that,
Distance = 20.0 m
Frictional force = 35.0 N
Angle = 25.0°
(A). We need to calculate the work done on the cart by the friction
Using formula of work done
Where, F = force
d = distance
Put the value into the formula
(B). The work done by the gravity is perpendicular to the direction of the motion
We need to calculate the work done on the cart by the gravitational force
Using formula of work done
Put the value into the formula
(C). We need to calculate the work done on the cart by the shopper
Using formula of work done
Put the value into the formula
(D). We need to calculate the force the shopper exerts
Using formula of force
Put the value into the formula
(E). We need to calculate the total work done on the cart
Using formula of work done
Put the value into the formula
Hence, (A). The work done on the cart by the friction is -700 J.
(B). The work done on the cart by the gravitational force is 0 J.
(C). The work done on the cart by the shopper is 700 J.
(D). The force the shopper exerts 38.61 N.
(E). The total work done on the cart is zero.
Answer:
754.6 m
Explanation:
The GPE (Gravitational potential energy) of an object with respect to the ground is given by
where
m is the mass of the object
g = 9.8 m/s^2 is the acceleration due to gravity
h is the heigth above the ground
Here we have
m = 12,400.05 kg is the mass
GPE = 91,700,000.00J is the GPE
Solving the formula for h, we find the heigth: