Aparcel of mass 10 kg rests on a lorry. when the lorry accelerates at 1.5 m/s^2,the parcel is just about to slide backwards. what is the co-efficient of friction between the parcel and the lorry?

The coefficient of static friction,  μ = 0.153

Explanation:

Given data,

The mass of the parcel, m = 20 kg

The acceleration of the lorry, a = 1.5 m/s²

The normal force acting on the parcel,

                                       f = m x g

                                         = 20 x 9.8

                                        = 196 N

The static frictional force acting on the lorry,

                                   F = 20 x 1.5

                                      = 30 N

The coefficient of static is

                     μ = F / f

                        = 30 / 196

                        = 0.153

Hence, the coefficient of static friction,  μ = 0.153

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm