Suppose you have to move a heavy crate of weight 875 N by sliding it along a horizontal concrete floor. You push the crate to th
e right with a horizontal force of magnitude 300 N, but friction prevents the crate from sliding.Part ADraw a free-body diagram of the crate.Part BWhat is the magnitude Fp of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficient of static friction s between the crate and the floor be 0.56 and that of kinetic friction, k, be 0.47.
1 answer:
A) See figure in attachment
There are 4 forces acting on the crate:
- The horizontal force F, pushing the crate to the right
- The frictional force , acting in the opposite direction (to the left)
- The weight of the crate, , acting downward (with m being the mass of the crate and g the acceleration due to gravity)
- The normal reaction of the floor agains the crate, N, acting upward, and of same magnitude of the weight
The crate is in equilibrium (it is not moving), this means that the forces along the horizontal direction and along the vertical direction are balanced, so:
B) 490 N
In order to make the crate start sliding along the floow, the horizontal push must overcome the maximum force of static friction, which is given by
where
is the coefficient of static friction
N is the normal reaction
The normal reaction is equal to the weight of the crate, so
and so, the maximum force of static friction is
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(a)
The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):
The initial kinetic energy is:
where m = 58.0 kg is the mass of the projectile and is the initial speed. Substituting,
The initial potential energy is given by
where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,
So, the initial mechanical energy is
(b)
We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.
The kinetic energy is
while the potential energy is
So, the mechanical energy is
And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:
And the work is negative because air friction is opposite to the direction of motion of the projectile.
(c) 88.1 m/s
The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:
When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:
E = K
The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:
And this is only kinetic energy:
So, we can solve to find the final speed: