FORCE AND LAWS OF MOTION

Solution: m1 = F/a1 ; andm2 = F/a2 . Here, a1 = 10 m s-2;a2 = 20 m s-2 and F = 5 N.Thus, m1 = 5 N/10 m s-2 = 0.50 kg; andm2 = 5 N/20 m s-2 = 0.25 kg.If the two masses were tied together,the total mass, m would be m = …

A force of 5 N gives a mass m1, an acceleration of 10 m s–2 and a mass m2 , an acceleration of 20 m s-2 . What acceleration would it give if both the masses were tied together? Read More »

Solution:The initial velocity of the motorcaru = 108 km/h= 108 × 1000 m/(60 × 60 s) = 30 m s-1and the final velocity of the motor carv = 0 m s-1.The total mass of the motorcar along with its passengers = 1000 kg and thetime taken to stop the motorcar, t = 4 s. we …

A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg. Read More »

Which would require agreater force –– accelerating a 2 kg massat 5 m s–2 or a 4 kg mass at 2 m s-2 ? Thus, F1 = m1 a1 = 2 kg × 5 m s-2 = 10 N; and F2 = m2 a2 = 4 kg × 2 m s-2 = 8 N.⇒ F1 …

Which would require a greater force –– accelerating a 2 kg mass at 5 m s–2 or a 4 kg mass at 2 m s-2 ? Read More »

Solution:We have been given that u = 3 m s–1 and v = 7 m s-1 , t = 2 s and m = 5 kg. we have,F = m ( v – u ) / tSubstitution of values in this relation givesF = 5 kg ( 7 m s-1 – 3 m s-1) / …

A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m s–1 to 7 m s-1. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object? Read More »