Work done W = F x d Only when there is distance moved in the direction of the force , there is work done. (i.e. look for the distance parallel to the force)

A 10 kg box is initially at rest on a rough horizontal surface and a horizontal force is then applied to it. The force varies as shown in the diagram. a) If the box moves with an acceleration of 2m/s2 in the first 15 s, what is the frictional force between the surfaces?   Resultant force on box F = ma Applied force – frictional force = ma 40 – frictional force  = 10 x 2                           frictional force = 20 N b) How does the velocity change during the next 5 s? Next 5 s after time 15 s, there is no applied force. Hence the resultant force acting on box isonly frictional force 20N (in direction opposite to motion of box). F = ma-20 = 10 x aa  = – 2 m/s2 The velocity of box at time 15s is required. Hence when t = 0s, box accelerates at 2m/s2 from rest, a = (v – u) / t 2 = (v – 0) /15 v = 30 m/s (Hence the final velocity of the box at time 15s is 30 m/s, just before applied force of 40 N is released) Find velocity of box at end of next 5s (time = 20s): a = (v – u) /t– 2  = (v – 30) / 5v = 20 m/s.    Therefore, velocity decreases at a constant rate of -2 m/s2 from  30 m/s to 20 m/s in next 5s. c) Plot a velocity-time graph and calculate the distance travelled by the box in the 30 s.

A uniform capillary tube closed at one end, contained air trapped by a thread of mercury 85 mm long. When the tube was held horizontal, the length of the air column was 50 mm. When it was held vertically with the closed end downwards, the length was 45 mm. Find the atmospheric pressure in Pa. (Density of mercury = 14 x 10^3 kg/m3 ) Solutions:

Solutions: Option A

A U-tube with 30 cm of water is set up vertically and 12 cm of olive oil is poured carefully into the left-hand limb of the tube as shown below. It is observed that water will be pushed up in the other limb. The density of water and olive oil are 1000 kg m-3 and 920 kg m-3 respectively. Given A is at same level as the boundary between olive oil and water. a) What is the length of water column above point A? b) A further 6 cm of olive oil is added into the left-hand limb. How much further will the water level rise in the right limb? Solutions: View video tutorial for part (b) a) Poil = Pwater above A ρgh = ρgh920 x 10 x 12 = 1000 x 10 x h h = 11.04 cm b)         Pressure of water = pressure of olive oil 1000 x 10 x (2x + 11.04)/100  = 920 x 10 x 18/100 x = 2.76 cm

Answer: Option A Refer to the video tutorial below for explanation.