Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
Answer: 2.5 m/s and 6.25 m
Explanation:
u = 0
a = 0.5 m/s²
t = 5 s
v = u + at
= 0 + 0.5 × 5
= <u>2.5 m/s</u>
s = ut + 1/2 at²
= 1/2 × 2.5 × 5
=<u> 6.25 m</u>
Answer:
296 N
Explanation:
Draw a free body diagram. The box has two forces on it: tension up and weight down.
Apply Newton's second law:
∑F = ma
T − mg = ma
T = m (g + a)
Given m = 196 N / 9.8 m/s² = 20 kg, and a = +5 m/s²:
T = (20 kg) (9.8 m/s² + 5 m/s²)
T = 296 N
In kynematics you describe the motion of particles using vectors and their change in time. You define a position vector r for a particle, and then define velocity v and acceleration a as
In dynamics Newton's laws predict the acceleration for a given force. Knowing the acceleration, and the kynematical relations defines above, you can solve for the position as a function of time: r(t)
