Given, mass of block,
m=1kg Acceleration due to gravity,
g=9.8ms−2 Inclination,
θ=30∘ Electric field,
E=200N∕C Coefficient of friction,
µ=0.2 Charge,
q=5mC=5×10−3C Let friction force,
f=µN where,
N be the normal reaction.
Since, net force is zero along the perpendicular direction of incline.
Therefore, force along
Y-axis will be zero.
⇒‌‌N‌‌=mg‌cos‌30∘+qE‌sin‌30∘ ⇒‌‌N‌‌=1×9.8×√3∕2+5×10−3×200×1∕2 ‌‌=8.49+0.5 ‌‌=8.99N∼eq9N ∴‌‌f=µN=‌×9=‌=1.8N Now, total force along the plane of incline,
‌mg‌sin‌30∘−f−qE‌cos‌30∘=ma‌ ⇒‌1×9.8×1∕2−1.8−5×10−3×200√3∕2‌=a ⇒‌5−1.8−1.732∕2‌=a ⇒‌a=2.34ms−2 Since, initial velocity of body,
u=0ms−1 and distance along incline,
s=h∕sin‌30∘=1∕sin‌30∘=2 By using second equation of motion,
‌s‌‌=ut+1∕2at2 ⇒‌2‌‌=0+1∕2×2.34×t2 ⇒‌t2‌‌=‌ ⇒‌t=‌