A car is travelling at 5 m/s up a gradient of 1 in 20. The car weight 6 tonnes and the friction is 48 kg wt. The work done per second in maintaining the motion of car up the gradient is $(g=10ms^{-2}$)

Given, speed of a car $v=5 \mathrm{~m} / \mathrm{s}$

gradient of 1 in 20, so, $\sin \theta=\frac{1}{20}$

weight of $\operatorname{car} m=6$ tones

$=6 \times 1000 \mathrm{~kg}$

friction force $f=48 \mathrm{~kg}-w t=(48 \times 10) \mathrm{N}$

we know when a block going rep, then

total force $F=m g \sin \theta+f$

$\begin{aligned} \Rightarrow F & =\left(6 \times 1000 \times 10 \times \frac{1}{20}+480\right) \\ & =3480 \mathrm{~N} \end{aligned}$

we know work done per secon $=$ powers $=F \cdot v$

$\begin{aligned} \Rightarrow \frac{d \omega}{d t}=P=F U & =(3480 \times 5) \text { watt } \\ & =17400 \mathrm{watt} \end{aligned}$