Question

A cart of mass 20 kg at rest is to be dragged at a speed of 18 km/hr.If the co-efficient of friction between the cart and the ground is 0.1,what is the mimimum force required to drag the cart to a distance of 10 m ? (take g= 10m/s)

Answer 1

Given, v = 18 km/h = 18×1000 3600=5 m/s2as=v2−u2⇒a=v2−u22s=2520=1.25 m/s2
where a is the acceleration of the cart

The minimum force required to drag the cart to distance 10 m = μmg+ma
F=μmg+ma⇒F=0.1×20×10+10×1.25 =32.5 N

Answer 2

Answer:45N

Explanation:Intial velocity, u=0 m/s

Final velocity, v=18km/h =5m/s

V2 - u2 = 2as

a =(v2 - U2) /2s

a =(25-0) /2×10

a = 25/20

a = 1.25 m/s^-2

The minimum force required to drag the cart to the distance 10 m

F = umg + ma

F = (0.1 × 20 × 10) +( 20 × 1.25)

F = 20 +25

F = 45N