Option (i) is correct because block P experiences an unbalanced net force of 1 N towards the right, while block Q moves at constant velocity, meaning the net external force acting upon Q is zero.
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For statement (i) the velocity will remain the same. For statement (ii) the magnitude of velocity will increase. For statement (iii) the magnitude of velocity will decrease due to opposing acceleration.
The frictional force exerted by the floor is equal in magnitude to F and acts in the direction opposite to motion. Net force is zero because the table moves with a constant velocity.
The impact velocity is 14.14 m s-1, derived from square root of two g h. Total kinetic energy equals 150 J, which divided by 3000 N resistive force gives a sand depression depth of 0.05 m.
Reading Fig. 7.39, total energy is 30 J. At P, potential energy is 20 J giving velocity 6.32 m s-1. At Q, potential energy is 30 J giving velocity 0 m s-1. Point R cannot be reached since it requires ...