A grocery cart with a mass of 15 kg is being pushed at constant speed up a 12∘ ramp by a force FP which acts at an angle of 17∘
below the horizontal.
a. Find the work done by the force F⃗ P on the cart if the ramp is 6.5 m long.
b. Find the work done by the force mg⃗ on the cart.
a. Find the work done by the force F⃗ P on the cart if the ramp is 6.5 m long.
b. Find the work done by the force mg⃗ on the cart.
2 answers:
Answer:
(a)The work done by the force Fp is 198.6J
(b) The work done by the weight mg is -198.6J
The net force acting on the cart is zero. This is because the cart is moving with a constant velocity and by newton's first law the the net force on the cart is equal to zero.
Fp was calculated to be equal to 34.94 N.
Explanation:
In order to solve this kind of problems successfully, the best approach is to resolve all forces acting on the cart parallel and perpendicular to the ramp surface. So that the x-axis is parallel to the ramp surface and the y axis is perpendicular to the ramp surface.
Fpx = FpCos 29°
Fpy = FpSin 29°
Wx = mg sin12°
Wy = mg cos 12°
Summation Fx = 0
And Summation Fy = 0
The full solution can be found below in the attachment.
Thank you for reading and I hope this is helpful to you.
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Answer:
0.001 s
Explanation:
The force applied on an object is equal to the rate of change of momentum of the object:
where
F is the force applied
is the change in momentum
is the time interval
The change in momentum can be written as
where
m is the mass
v is the final velocity
u is the initial velocity
So the original equation can be written as
In this problem:
m = 5 kg is the mass of the fist
u = 9 m/s is the initial velocity
v = 0 is the final velocity
F = -45,000 N is the force applied (negative because its direction is opposite to the motion)
Therefore, we can re-arrange the equation to solve for the time: