A 2,000 kg car traveling at 30.0 mi/h skids to a stop in 60.0 meters. The force

Question

A 2,000 kg car traveling at 30.0 mi/h skids to a stop in 60.0 meters. The force of friction during the skid is (1.0 mi/h = 0.447 m/s) A. 4,500 N. B. 4,200 N. C. 3,750 N. D. 3,000 N. E. 2,400 N. 21. The explosion in a cannon exerts an average force of 30,000 N for 2.00 meters. The velocity of a 2.00 kg projectile shot from the cannon is A. 198 m/s. B. 245 m/s. C. 302 m/s. D. 354 m/s. E. 412 m/s. 22. A 20.0 kg box slides up a 12.0 m long incline at an angle of 30.0 degrees with the horizontal. A force of 150 N is applied to the box to pull it up the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then the increase in the kinetic energy of the box is A. 330 J. B. 390 J. C. 420 J. D 481 J. E 597 J.

Explanation / Answer

here,

18)

speed of plane , v = 750 km/h

v = 208.33 m/s

let the radius of the arc be r

m * g = m * v^2/r

9.8 = 208.33^2 /r

r = 4429 m = c) 4.4 km

20)

initial speed , u = 30 mi/h

u = 13.41 m/s

s = 60 m

let the acccelration be a

using third equation of motion

v^2 - u^2 = 2 * a * s

0 - 13.41^2 = 2 * a * 60

a = 1.5 m/s^2

force applied , F = m * a

F = 2000 * 1.5 = D) 3000 N

22)

mass , m = 20 kg

l = 12 m

theta = 30 degree

F = 150 N

uk = 0.1

the accelration of box , a = net force /mass

a = ( F - m*g*sin(theta) - uk * m * g * cos(theta)) /m

a = ( 150 - 20 * 9.8 * sin(30) - 0.1 * 20 * 9.8 * cos(30)) / 20

a = 1.75 m/s^2

final velocity , v = sqrt(2 * a * l) = 6.48 m/s

the kinetic energy gained , KE = 0.5 * m * v^2

KE = 0.5 * 20 * 6.48^2

C) KE = 420 J

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