A block of mass 2 kg is moving with initial velocity 10 m/s enters on a rough horizontal surface. An extra force of 20 N is opposing the motion and trying to stop the block. The time after which the block will stop momentarily is (g = 10 m/s2)

Question

A block of mass 2 kg is moving with initial velocity 10 m/s enters on a rough horizontal surface. An extra force of 20 N is opposing the motion and trying to stop the block. The time after which the block will stop momentarily is (g = 10 m/s2) (A) 1 s (B) 0.5 s (C) 0.67 s (D) 0.33 s

Tardigrade Admin · Accepted Answer

Fnet 20+0.5× 20 = 30 N Reatardation a = (Fnet/m)=(30/2) = 15m/s2 t = (v/a)=(10/15)=0.67s

Q. A block of mass 2 kg is moving with initial velocity 10 m/s enters on a rough horizontal surface. An extra force of 20 N is opposing the motion and trying to stop the block. The time after which the block will stop momentarily is ( $g = 10 m / s^{2}$ )

1 s

0.5 s

0.67 s

0.33 s

$F_{n e t} 20 + 0.5 \times 20 = 30 N$
Reatardation a = $\frac{F _{n e t}}{m} = \frac{30}{2} = 15 m / s^{2}$
$t = \frac{v}{a} = \frac{10}{15} = 0.67 s$

Q. A block of mass 2 kg is moving with initial velocity 10 m/s enters on a rough horizontal surface. An extra force of 20 N is opposing the motion and trying to stop the block. The time after which the block will stop momentarily is (g=10m/s2)

1 s

0.5 s

0.67 s

0.33 s

Fnet​20+0.5×20=30N Reatardation a = mFnet​​=230​=15m/s2 t=av​=1510​=0.67s

Q. A block of mass 2 kg is moving with initial velocity 10 m/s enters on a rough horizontal surface. An extra force of 20 N is opposing the motion and trying to stop the block. The time after which the block will stop momentarily is ( $g = 10 m / s^{2}$ )

$F_{n e t} 20 + 0.5 \times 20 = 30 N$
Reatardation a = $\frac{F _{n e t}}{m} = \frac{30}{2} = 15 m / s^{2}$
$t = \frac{v}{a} = \frac{10}{15} = 0.67 s$