Question

QK3 An object of mass 5kg starts from rest moves with velocity 5 m/s in time interval 10s, calculate the
distance travelled and also calculate the force applied by the object.​

Answer 1

Answer:

force is 3.75N

Step-by-step explanation:

Apply equation of kinematic

v=u+at

Acceleration, a=

t

v−u

=

20

25−10

=0.75ms

−2

Force= mass x acceleration

F=5×0.75=3.75N

Hence, force is 3.75N

Answer 2

Distance = 25m

Force = 2.5N

Step-by-step explanation:

Given -

• Mass(m) = 5kg
• Initial velocity (u) = 0m/s
• Final velocity (v) = 5m/s
• Time (t) = 10 seconds

Solution -

Acceleration -

• a= v-u/t
• 5-0/10
• 5/10
• 0.5m/s²

Acceleration is 0.5m/s²

Distance(s) -

• ut+1/2at²
• (0×10)+1/2(0.5)(10)²
• 1/2 (0.5)(100)
• 50/2
• 25m

Distance is 25m

Force-

• ma
• 5×(0.5)
• 2.5N

Force is 2.5N