Question

A 850-kg pickup decelerates to rest from a speed of 75 km/h in a distance of 145 m. Presume the pickup is initially traveling in the positive direction.
a) If the brakes are the only thing making the pickup come to a stop, calculate the fore in newtons, in a component along the direction of motion of the pickup, that the brakes apply on the pickup Fb?
b) Suppose instead of braking that the pickup hits a concrete abutment at full speed and is brought to a stop in 2.00 m. calculate the force in newtons exerted on the pickup in this case?
c) what is the ratio of the force on the pickup from the concrete to the braking force?

Answer 1

here,

mass , m = 850 kg

initial speed , u = 75 km/h = 20.83 m/s

a)

the distance travelled before stopping , s = 145 m

let the acceleration of mass be a

using third equation of motion

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

0^2 - 20.83^2 = 2 * a * 145

solving for a

a = - 1.5 m/s^2

the force applied by the brakes , F = m * a

F = 850 * (-1.5) N = - 1272.15 N

b)

the distance travelled before stopping , s' = 2 m

let the acceleration of mass be a'

using third equation of motion

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

0^2 - 20.83^2 = 2 * a' * 2

solving for a'

a' = - 108.47 m/s^2

the force applied by the brakes , F' = m * a'

F' = 850 * (-108.47) N = - 9.22 * 10^4 N

c)

the ratio of the force on the pickup from the concrete to the braking force , R = F' /F

R = 9.22 * 10^4 /1272.15

R = 72.48