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Answers
Given :
A contact force acts on an object of mass 5kg for a duration 2minutes .
it increases the objects velocity from 18 m/s to 30 m/s.
To Find :
The magnitude of the force applied = ?
Solution :
Initial velocity (u) = 18 m/s
Final velocity (v) = 30 m/s
mass (m) = 5 kg
Time (t) = 2 minutes
First of all convert the time taken from minutes to seconds :
→ Time = 2 minutes
→ Time = 2 × 60 seconds
→ Time = 120 seconds
Hence,a contact force acts on an object of mass 5kg for a duration 120 seconds.
Finding the acceleration of the object :
→ Acceleration = (Final velocity - Initial velocity) ÷ Time
→ Acceleration = (30 - 18) ÷ 120
→ Acceleration = 12 ÷ 120
→ Acceleration = 0.1 m/s²
Hence,the acceleration of the object is 0.1 m/s².
Now, let's find the magnitude of the force applied :
→ Force = mass × acceleration
→ Force = 5 × 0.1
→ Force = 0.5 N
Hence,the magnitude of the force applied is 0.5 N.
Answer:
Given:
A motor cyclist ( to be treated as a point mass) is to undertake horizontal circles inside the cylindrical wall of a well of inner radius 8m. (use g = 10 m/s²)
To find:
Find the coefficient of static friction between the tyres and the well when the minimum speed to perform this stunt is 12m/s
Solution:
From given, we have,
The speed of the motor cyclist = v = 12 m/s^2
The radius of the cylindrical wall of a well = r = 8 m
We know that the necessary centripetal force is provided by the friction between the road and the tyres, so we have,
F = mv²/r = μmg
μ = v²/rg
μ = 12² / (8 × 10)
μ = 144 / 80
μ = 1.8
Therefore, the coefficient of static friction between the tyres and the well is 1.8