Question


At the instant a race began, a 89.1 kg sprinter exerted a force of 710 N on the starting block in the direction 110 degrees measured clockwise from the positive x-axis.
(a) What was the horizontal acceleration of the sprinter?

a_x =

(b) If the force was exerted for 0.7 s, what is the velocity of the sprinter?

v_{final} =
a. Fe = 710N.[11o] = Exerted force.
Xe = 710*cos11 = 697.0 N. = Hor.
component of exerted force.
a = Xe/m = 697/89.1 = 7.82 m/s^2.

b. V = Vo + a*t = 0 + 7.82*0.7=5.48 m/s.

NOTE: I used 11o for the given angle,
because 110o is not possible.

Related Questions

ok so I have a question At the instatn a race began, a 57-kg sprinter was found to exert a force... At the instant a race began, a 74.0 kg sprinter exerted a force of 850 N on the starting block at a... At the instant a race began, a 60-kg sprinter exerted a force of 669-N on the starting block at a 24... A 67.5 kg sprinter exerts a force of 775 N on the starting block which makes a 20 degree angle to th...