UNIT-1: Chapter-2
FORCES AND SHAPE
Solution: Step 1: Choose a scale of 1 cm to represent 5 N. Arrows OA and OB represent the 15 N and 20 N forces.
Step 2: Complete the parallelogram OACB.
Step 3: The resultant is 5 cm in length, i.e. resultant = 25 N, acting at an angle of 37⁰ to the horizontal.
WORK EXAMPLE 3.3
A boy pushes a box of mass 20 kg with a force of 50 N. What is the acceleration of the box? (Assume no friction.)
Solution: Given: mass m = 20 kg
force F = 50 N
From Newton’s Second Law,
F = ma Where a = acceleration of the box
Therefore a = = = 2.5 ms-2
WORK EXAMPLE 3.4
A car of mass 1000 kg acceleration from rest to 20 ms-1 in a time of 5s. Calculate the forward thrust of the car. (Assume no friction.)
Solution:
Given: mass, m = 1000 kg
Initial speed, u = 0 ms-1
Final speed, v = 20 ms-1
Time, = 5 s
From Newton’s Second Law,
Forward thrust, F= ma where a= acceleration produced
but a = v-u/t
= 20-0/5
= 4 ms-2
Therefore, F = ma = (1000)(4)= 4000 N
WORK EXAMPLE 3.5
A mass of weight W equal to 6.0 N hangs on the end of a string which is pulled sideways by a force F so that the string makes an angle of 30⁰ with the vertical as shown in Figure 3.29.
The tension T in the string is 7.0 N. By means of a force parallelogram, determine the force F.
Solution: Since the mass is stationary, the forces W, T and F have no resultant.
From the force parallelogram, F has a length of 3.5 cm. Hence,
F = 3.5 N
WORK EXAMPLE 3.6
The sliding frictional force between a box of mass 4 kg and the floor is 15 N. It is pushed across the floor with a constant force such that it accelerates at 0.8 ms-2.
(a) What is the force applied to the box?
.jpg)
(b) If this same force calculated in (a) is applied to the box which is now placed on a frictionless floor, what is the new acceleration produced?
Solution: (a) acceleration a = 0.8 ms-2
By Newton’s Second Law,
resultant (net) force = mass × acceleration
F – f = ma
F = f + ma
= 15 (4) (0.8)
= 18.2 N which is the required constant force.
(b) For frictionless floor, f = 0.
Hence, F = ma’ where a’ = new acceleration
a’ = F/m
= 18.2/4
= 4.6 ms-2
Assistant Teacher (Senior section), O and A level Physics Teacher, Bangladesh International School and College, Dhaka.
কলি
