a) a 100m sprinter was told that the coefficient of static friction is 0.800 between the sole of her shoes and the track surface. if her mass is 60 kn,what maximum acceleration can she achieve?
b) a box of books weighing 300N is pushed across the floor by a force of 400N downwards at an angle of 37 below the horizontal. if the coefficient of kinetic friction between the box and the floor is 0.47, how long will it take to push the box 4m across the hall?
b) a box of books weighing 300N is pushed across the floor by a force of 400N downwards at an angle of 37 below the horizontal. if the coefficient of kinetic friction between the box and the floor is 0.47, how long will it take to push the box 4m across the hall?
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a) Newton's Third law: action and reaction. The sprinter presses against the ground and the reaction force to the static friction force allows her to accelerate.
f = 60kg * 9.8 m/s^2 * 0.8 = 470.4 N.
This force 470.4N accelerates the sprinter:
F = ma, a = F/m = 470.4 N/ 60kg = 7.84 m/s^2
b) The force due to kinetic friction is 300N * 0.47 = 141. The applied force against this kinetic friction is 400 N cos 37 = 319.45
The net force is 319.45 - 141 = 178.45 N
F = ma, and a = F/m.
Here, the books weigh 300N, their mass would be 300N / 9.8 m/s^2 = 30.61 kg
Therefore, a = 178.45 N / 30.61 kg = 5.83 m/s^2
Assuming that the box was not moving initially, the distance traveled by the box with this acceleration is,
S = 1/2 * a t^2
Now set S be 4 m and solve for t:
t^2 = 1.372
so
t = 1.17 secs
f = 60kg * 9.8 m/s^2 * 0.8 = 470.4 N.
This force 470.4N accelerates the sprinter:
F = ma, a = F/m = 470.4 N/ 60kg = 7.84 m/s^2
b) The force due to kinetic friction is 300N * 0.47 = 141. The applied force against this kinetic friction is 400 N cos 37 = 319.45
The net force is 319.45 - 141 = 178.45 N
F = ma, and a = F/m.
Here, the books weigh 300N, their mass would be 300N / 9.8 m/s^2 = 30.61 kg
Therefore, a = 178.45 N / 30.61 kg = 5.83 m/s^2
Assuming that the box was not moving initially, the distance traveled by the box with this acceleration is,
S = 1/2 * a t^2
Now set S be 4 m and solve for t:
t^2 = 1.372
so
t = 1.17 secs
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a)max value of friction
f = uN = umg
so man can exert maximum force equal to friction
f = Fmax
ma = f
umg = ma
a = ug = 0.800*9.8 = 7.84m/s^2
b) mg = 300 , m = 300/9.8 = let F is the force applied
in y-direction
N = mg+Fsin37
also
f = uN = u(mg+Fsin37) ....(1)
in X direction
Fcos37- f = ma
Fcos37 - umg - uFsin37 = ma
F(cos37 - usin37) - umg = ma
a = 2.16m/s^2
s= 0+1/2at^2
8 = 2.16*t^2
t = 1.9sec
f = uN = umg
so man can exert maximum force equal to friction
f = Fmax
ma = f
umg = ma
a = ug = 0.800*9.8 = 7.84m/s^2
b) mg = 300 , m = 300/9.8 = let F is the force applied
in y-direction
N = mg+Fsin37
also
f = uN = u(mg+Fsin37) ....(1)
in X direction
Fcos37- f = ma
Fcos37 - umg - uFsin37 = ma
F(cos37 - usin37) - umg = ma
a = 2.16m/s^2
s= 0+1/2at^2
8 = 2.16*t^2
t = 1.9sec