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Question

3 questions

#A force of 5kgf acts on a mass of 2kg for 5 sec. Cal:- Acceleration produced, The distance moved by the body, The velocity of the body after 5 sec.

#A constant force of 98N acts in a body for 10sec and produces in I a velocity of 35m/s.Find mass of the body.

#A force of 0.01N acts on a body of mass 0.5kg. How much distance will the body travel in 10sec. Assuming friction to be absent.

pls show the steps and show me if u hav used a formula

pls answer even one

Answer 1

first question
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1 kg-f = 9.8 N
F = 5 * 9.8 = 49 N
m = 2 kg
t = 5 s

a = F/m = 49/2 = 24.5 m/s^2
acceleration is 24.5 m/s^2

S = ut + (1/2)at^2
S = (1/2) * 24.5 * 5^2
S = 306.25 m
Distance travelled is 306.25 m

v = u + at
v = at
v = 24.5 * 5 = 122.5 m/s
velocity is 122.5 m/s

second question
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F = 98 N
t = 10 s
v = 35 m/s

m = F/a

v = u + at
v = at
a = v/t
a = 35/10 = 3.5 m/s^2

therefore mass m = 98 / 3.5 = 28 kg

last question
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F = ma
m = 0.5 kg
F = 0.01 N
t = 10 s

a = F/m = 0.02 m/s^2
S = ut + (1/2)at^2
u = 0;
S = (1/2)at^2
S = (1/2) * 0.02 * (10)^2
S = 1 m
Distance travelled is 1 m

Answer 2

fully agree with finite_st...