Another PHYSICS question :)

Question

I'm in need of your help guys.

So, here are the problems:

(1)

A person with a black belt in karate has a fist that has a mass of 0.70kg. Starting from rest, this fist attains a velocity of 8.0 m/s in 0.15 seconds. What is the magnitude of the average net force applied to the fist to achieve this level of performance?

(2)

A 292-kg motorcycle is accelerating up along a ramp that is inclined 30 degrees above the horizontal. The propulsion force pushing the motorcycle up the ramp is 3150N, and air resistance produces a 250N that opposes the motion. Find the magnitude of the motorcycle's acceleration.

First one to answer will get 10 points. :)

Thanks for reading this.

Answer 1

I thought you've said "Another one question.."  singular, but you have two....Okay let's see

(1)

0 to 8 m/s in 0.15 sec not too bad

F= m (v2- V1) / (t2-t1)

F= 0.70 x ( 8/0.15)

F=37.3 N ( this black belt may need a bit of training)

(2) What is the force of the gravity component Fg that is pushing the bike backward?

Fg= mg sin(30)

m= 292-kg

If the force proppeling the bike is F1 =3150 N

and wind resistance is Fw= 250 N then

Ft= F1 - Fw - Fg

a= Ft/m

a= ( 3150 - 250 - mg sin(30) )/m

a=(2900 - 292 x 9.81 x 0.5)/ 292=5.0m/s^2

Answer 2

1) f=ma (force = mass * acceleration)

m = 0.7

v = at, a= v/t (accel = time/mass), so a = 8 / 0.15 = 53.33333

f = 0.7 * 53.3333 = 37.3333 newtons

2) f = (3150-250) = 2900N

f=ma, a=f/m = 2900/292 = 9.93 m/s2 (meter per sec squared)

Note, answer #2 depends on the force pushing the moter cycle to be inclined as well. If the force is parallel to the ground (vs. the incline), then you have to take the cosine of 30, multiply by 3150, subtract 250, then divide by 292, for a net of 8.49 m/s2

Also, I did not consider gravity (i read the question as netting out the force moving it upward). The question is a bit vague, I believe.