Respuesta :
Answer:
Ff = 0.01107 N - Frictional force
P = 0.02491 N - Applied force
Explanation:
Given:-
- The mass of the skateboard, m = 0.50 kg
- The first part of journey was for distance, s1 = 1.0 m
- The time duration of first part of journey, t1 = 8.5 s
- The second part of the journey was for distance, s2 = 1.25 m
- The initial and final conditions of the skateboard are at rest.
Find:-
Calculate the force applied to the skateboard, and constant frictional force opposing its motion.
Solution:-
- The first part of the journey had two forces acting on the skateboard over the distance "s1" and time duration "t1".
- The applied force " P " makes the skateboard move and the opposing frictional force " Ff " opposes the motion.
- Assuming the linear acceleration " a " of the skateboard to remain constant.
- Apply the second kinematic equation of motion:
s1 = u*t1 + 0.5*a*t1^2
Where,
u: The initial velocity. ( skateboard is at rest; hence, 0 )
s1 = 0.5*a*t1^2
a = 2*s1 / t1^2
a = 2*1 / 8.5^2
a = 0.02768 m/s^2
- Similarly, the second part of the journey devoid of the applied force " P " and only the frictional force "Ff" acts on the block slowing the skateboard to rest.
- Apply the first equation of motion to determine the speed "v1" as soon as the force "P" is removed or the start of the second journey.
v1 = u + a*t
v1 = 0 + 0.02768*8.5
v1 = 0.23528 m/s
- The block slows down from the speed of v = 0.23528 to zero over the distance of s2 = 1.25. The linear acceleration for the second part of the journey ( a2 ) can be determined from the 3rd equation of motion:
v2^2 = v1^2 + 2*a2*s2
Where,
v2: The final velocity of second journey ( rest ) = 0
0 = 0.23528^2 + 2*a2*1.25
a2 = - 0.23528^2 / 2*1.25
a2 = - 0.02214 m/s^2
- The second part of the journey the skateboard decelerates to rest with linear acceleration of "a2".
- Apply the Newton's second law of motion for the second part of the journey:
F_net = m*a2
Where,
F_net = - Ff ( Frictional force )
-Ff = 0.5*( -0.02214 )
Ff = 0.01107 N ... Answer
- Apply the Newton's second law of motion for the first part of the journey:
F_net = m*a
Where,
F_net = P - Ff ( Both forces act )
P - Ff = 0.5*( 0.02768 )
P = 0.01384 + 0.01107
P = 0.02491 N .... Answer
