Question

Please help me in this one :
A cart of mass 20 kg at rest is to be dragged at a speed of 18 km/h. If the coefficient of friction between the cart and the ground is 0.1, what is the minimum force required to drag the cart to a distance of 10m. (g = 10m/s²) ​

Answer 1

Answer:

A cart of mass 20 kg at rest is to be dragged at a speed of 18 km/h. If the coefficient of friction between the cart and the ground is 0.1, The minimum force required to drag the cart to a distance of 10m is 45 N.

Explanation:

Given,

Mass of the cart (m)                     =    20 kg

Acceleration due to gravity (g)    =    10m/s²​

speed (v)                  =  18km/hr = $18\times{\frac{5}{18} } = 5 m/s$

Initial speed (u)         = 0

Coefficient of friction ( μ ) = 0.1

When the cart starts to move by applying a force F,  the frictional force( $f_{r}}$) acts in the opposite direction which can  be written as,  

 $f_{r} =$ μ N

where N - normal force, N= mg = 20 $\times$ 10 =200N

$f_{r} =$ $0.1 \times 200 = 20 N$....(1)

so, the net force $F_{net}$  is given by,

$F_{net} = F - f_{r}$ ....(2)

Also, we have Newton's law of motion as,

$F_{net} =ma$

$F - f_{r} = ma\\\\F = ma + f_{r}$ ...(3)

To find the value of a, consider the equation,

$v^{2}=u^{2}+2 a s$  

$v = 5m/s }$

$u = 0$

s = 10m

subtituting the values,

$(5)^{2}=0+2 \times a \times 10$

$a = \frac{25}{2\times10} = \frac{5}{4} m/s^{2}$

Equation (3) becomes,

$F = (20\times \frac{5}{4}) + 20$

   F  = 45 N