A0.060 kg ball hits the ground with a speed of -32m/s. the ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 n force on the ball. what is the magnitude of the velocity after it hits the ground?

a. 9.3m/sb.12m/sc.41m/sd.73m/s

A:

Impulse is given by:

FΔt = Δp

We know the mass and velocity and we need the final velocity so lets break this down and isolate

FΔt = mv(final) - mv(initial)

Factor out m

FΔt = m(v(final) - v(initial))

FΔt / m = v(final) - v(initial)

Add v initial

(FΔt / m) + v(initial) = v(final)

(55x45ms / 0.060) - 32 = v(final)

v(final) = +9.25 m/s or 9.3

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards