Problem 2: In the figure, the net extenmal force on the 24 kg mower is known to

Question

Problem 2: In the figure, the net extenmal force on the 24 kg mower is known to be 51 N Randomized Variables f 29 N v=1.21m/s Ctheexpertta.com Part (a) If the force of friction opposing the motion is 29 N, what force F (in newtons) is the person exerting on the mower? Numeric : A numeric value is expected and not an expression. Part (b) Suppose the mower is moving at I.2 m/s when the force F is removed. How far will the mower go before stopping in m? Numerie : A numeric value is expected and not an expression.

Explanation / Answer

a) Fnet = F_person - f

==> F_person = Fnet + f

= 51 + 29

= 70 N

b) vi = 1.2 m/s

a = -f/m

= -29/24

= -1.208 m/s^2

let d is the distance travelled by the mower before coming to stop.

use, vf^2 -vi^2 = 2*a*d

==> d = (vf^2 - vi^2)/(2*a)

= (0^2 - 1.2^2)/(2*(-1.208))

= 0.596 m

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