Question 21:
easyAssertion (A): The apparent weight of a person standing in a lift, which speed up is always greater than his true weight.
Reason (R): The gravity force due to earth always acts downwards.
If the lift speeds up *upwards*, apparent weight \(N = m(g+a)\), which is greater than true weight \(mg\). Under this interpretation, (A) is true. Gravity always acts downwards, so (R) is true. However, (R) does not explain the change in apparent weight due to acceleration.
Question 22:
easyWhich one of the following statements is incorrect?
(2018)
Friction force always opposes relative motion. Limiting static friction \( \text{f}_{\text{s,max}} \) is proportional to normal reaction \( \text{N} \), so \( \text{f}_{\text{s,max}} = \mu_{\text{s}}\text{N} \). Rolling friction is generally much smaller than sliding friction. The coefficient of friction \( \mu \) is a ratio of forces (friction force to normal force), thus it is a dimensionless quantity. Therefore, option (d) is incorrect.
Question 23:
easyA block of mass \(10\text{ kg}\) placed on rough horizontal surface having coefficient of friction \(\mu = 0.5\), if a horizontal force of \(100text{ N}\) acting on it then acceleration of the block will be:
(2002)
Calculate max friction \(f = \mu mg\). Compare with applied force (F). If (F > f), then (a = (F-f)/m). Here, (f = 0.5 \times 10 \times 10 = 50\text{ N}). So, (a = (100-50)/10 = 5\text{ m/s}^2).
Question 24:
easyOn the horizontal surface of a truck a block of mass \(1\text{ kg}\) is placed \((\mu = 0.6)\) and truck is moving with acceleration \(5\text{ m/s}^2\) then the frictional force on block will be:
(2001)
Required force for block to accelerate with truck is \(F = ma_{truck}\). Maximum static friction is \(f_{s,max} = \mu mg\). If \(F \leq f_{s,max}\), friction equals (F). Here, \(F = 1 \times 5 = 5\text{ N}\) and \(f_{s,max} = 0.6 \times 1 \times 10 = 6\text{ N}\). Since \(5 \leq 6\), the frictional force is \(5\text{ N}\).
Question 25:
easyConsider a car moving along a straight horizontal road with a speed of \(72\text{ km/h}\). If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance, in which the car can be stopped is (taking \(g = 10\text{ m/s}^2)\):
(1992)
Convert speed to m/s: \(v = 72 \times 5/18 = 20\text{ m/s}\). Deceleration due to friction is \(a = \mu_s g = 0.5 \times 10 = 5\text{ m/s}^2\). Using \(v^2 = u^2 + 2as\), \(0 = (20)^2 - 2(5)s\). So, \(10s = 400\) and \(s = 40\text{ m}\).