Question

1) Little Timmy wants to measure how tall his house is. He doesn’t have a tape measure but does have a stopwatch. He recruits Big Bob to climb on his roof and drop a rock. He times the rocks fall and finds it to be 1.67 s. How tall is his house?

2) A large stuffed chicken is shot of the edge of a cliff horizontally with an initial velocity of 36.0 m/s. If the cliff is 42.2 m high, how far from the bottom of the cliff with the chicken land?



3) A football is kicked at a velocity of 22.0 m/s at an angle of 45o off the ground. How far away does the football land?



4) Taylor Swift is pushing a large monkey across the floor with a force of 345 N. If the monkey has a mass of 31.0 kg, and the coefficient of kinetic friction is 0.16, what is the acceleration of the monkey?



5) A large box of pocket sized elephants is being pushed across the floor with a force of 648 N that is 25o above the horizontal. The mass of the box is 75.0 kg and the coefficient of kinetic friction is 0.17, what is the net force on the box?



6) A table with a mass of 25.0 kg has the following items sitting on it: A lap giraffe (63 N), A blueberry pie scented candle (12 N), a small golden figurine of Chuck Norris (44 N), A light saber (24 N), a wet towel (9 N), and a copy of Mario Kart for SNES (10 N). If the coefficient of static friction between the table and the floor is 0.22 how much force would be necessary to get the table moving?

Answer 1

1) 13.7 m

The motion of the rock is a free fall, with constant acceleration g=9.8 m/s^2 towards the ground, so the total distance it covers is given by the SUVAT equation:

`S=(1)/(2)gt^2`

where S is the height of the house, and t is the time the rock takes to reach the ground. Substituting t=1.67 s, we find:

`S=(1)/(2)(9.8 m/s^2)(1.67 s)^2=13.7 m`

2) 105.5 m

The motion of the stuffed chicken is a projectile motion, with a uniform horizontal motion (with constant velocity of v=36.0 m/s) and a vertical accelerated motion (with constant acceleration of g=9.8 m/s^2).

First of all, we can find the total time of the ball by considering the vertical motion only. We know the vertical distance covered, S=42.2 m, so the time of the fall is

`S=(1)/(2)gt^2\\t=\sqrt{(2S)/(g)}=\sqrt{(2(42.2 m))/(9.8 m/s^2)}=2.93 s`

And now we can consider the horizontal motion to find the horizontal distance covered by the stuffed chicken:

`d=vt=(36.0 m/s)(2.93 s)=105.5 m`

3) 49.4 m

Again, the motion of the ball is a projectile motion, with a horizontal motion and a vertical motion.

The range of a projectile launched from the ground can be found by using the formula:

`d=(v^2)/(g) sin 2 \theta`

where, in this case:

v = 22.0 m/s is the initial velocity

`\theta=45^(\circ)`

Substituting into the formula, we find

`d=((22.0 m/s)^2)/(9.8 m/s^2)(sin (2\cdot 45^(\circ)))=49.4 m`

4) 9.6 m/s^2

The frictional force acting on the monkey is given by:

`F_f = \mu mg=(0.16)(31.0 kg)(9.8 m/s^2)=48.6 N`

where
`\mu` is the coefficient of friction and m is the mass of the monkey.

We have two forces acting on the monkey: the push of F=345 N and the frictional force acting in the opposite direction. According to Newton's second law, the net force will be equal to the product between the monkey's mass and its acceleration, so we can find the acceleration:

`F-F_f=ma\\a=(F-F_f)/(m)=(345 N-48.6 N)/(31.0 kg)=9.6 m/s^2`

5) 462.3 N

The horizontal component of the pushing force is:

`F_x = F cos \theta = (648 N)(cos 25^(\circ))=587.3 N`

The frictional force, acting in the opposite direction, is

`F_f = \mu mg=(0.17)(75.0 kg)(9.8 m/s^2)=125.0 N`

where
`\mu` is the coefficient of friction and m is the mass of the box.

The net force on the box is therefore given by the net force on the horizontal direction:

`F_(net)=F_x -F_f=587.3 N -125.0 N=462.3 N`

6) 89.5 N

First of all we need to calculate the total weight of the table and the items above it.

The weight of the table is:

`W=mg=(25.0 kg)(9.8 m/s^2)=245 N`

So the total weight of the table and the items is

`W=245 N+63 N+12 N+44 N+24 N+9N+10N=407 N`

The force needed to get the table moving must be at least equal to the frictional force, which is equal to the product between the coefficient of friction and the weight of the all stuff:

`F=F_f = \mu W=(0.22)(407 N)=89.5 N`