An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field 200 ( N / C ) as shown in the figure. A body of mass 1 kg and charge 5 mC is allowed to slide down from rest at a height of 1 m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom. [ g =9.8 m / s 2, sin 30°=(1/2).; . cos 30°=(√3/2)]

Question

An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field 200 ( N / C ) as shown in the figure. A body of mass 1 kg and charge 5 mC is allowed to slide down from rest at a height of 1 m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom. [ g =9.8 m / s 2, sin 30°=(1/2).; . cos 30°=(√3/2)] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9982f8e517273a97ddbe7197ed245360-.png" /> (A) 0.92 s (B) 0.46 s (C) 2.3 s (D) 1.3 s

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7ac38a57bf1fe8558abbe7c83f551422-.png" /> here N =9.8 cos 30+1 sin 30 ≈ 9 N so a =(9.8 sin 30-1 cos 30-μ N /1) a =2.233 m / s 2 By S = ut +(1/2) at 2 =(1/2)(2.233) t 2 sin 30° t ≈ 1.3 sec

$0.92 s$

$0.46 s$

$2.3 s$

$1.3 s$

here $N = 9.8 cos 30 + 1 sin 30$
$\approx 9 N$
so $a = \frac{9.8 s i n 30 - 1 c o s 30 - μ N}{1}$
$a = 2.233 m / s^{2}$
By $S = u t + \frac{1}{2} a t^{2}$
$= \frac{1}{2} (2.233) t^{2}$
$sin 3 0^{\circ}$
$t \approx 1.3 sec$

0.92s

0.46s

2.3s

1.3s

here N=9.8cos30+1sin30 ≈9N so a=19.8sin30−1cos30−μN​ a=2.233m/s2 By S=ut+21​at2 =21​(2.233)t2 sin30∘ t≈1.3sec

$0.92 s$

$0.46 s$

$2.3 s$

$1.3 s$

here $N = 9.8 cos 30 + 1 sin 30$
$\approx 9 N$
so $a = \frac{9.8 s i n 30 - 1 c o s 30 - μ N}{1}$
$a = 2.233 m / s^{2}$
By $S = u t + \frac{1}{2} a t^{2}$
$= \frac{1}{2} (2.233) t^{2}$
$sin 3 0^{\circ}$
$t \approx 1.3 sec$