Question

3. A constant force acts on a 5.0 kg object and reduces its velocity from 7.0 m/s to 3.0 m/s in a time of 3.0 s. Find the force.

4. A 600-kg car is moving on a level road at 30 m/s. (a) How large a retarding force (assumed constant) is required to stop it in a distance of 70 m? (b) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction – there’s no sliding.

5. How much work is done against gravity in lifting a 3.0-kg object through a vertical distance of 40 cm?

Answer 1

@starfireandrobin

Answer 2

I'm sorry for leaving you hanging. The first part of the question, part 3, I know you have to use the force formula, which is F=ma.

Answer 3

Since F needs to be in newtons do i convert anything?

Answer 4

3. \[a= \frac{v-u}{t}\]where v=final velocity
u=initial velocity
& t=time taken

Finally F=ma

Answer 5

4. (i)\[v^2=u^2+2as\] where, v=final velocity=0 m/s
u= initial velocity=30 m/s
a= acceleration
s= distance traveled=70 m

After finding the acceleration from the above equation

retarding force = F=ma.
(ii)Retarding force=F=\(\mu_k N\)
where N is the normal force,

From here find coefficient of kinetic friction (\(\mu_k\)).

Answer 6

5. work done in lifting against gravity = \(-\Delta U\)=\(-mgh\)
where h is the final height of object

Answer 7

it can be solved using Momentum

f = m dv/dt

f dt = mdv << integrate


f delta(t) = m delta(v)

m = 5kg , v1 = 7 m/s , v2 = 3 m/s , delta(t)= 3 s

f * 3 = 5 ( 7-3)

f * 3 = 5 * 4

f = 20/3 N