Explanation:
By the second law of Newton we get the relation
F = ma
ANSWER:
F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.
STEP-BY-STEP EXPLANATION:
F(h) is Horizontal Force = 200 N
V is Speed = 2.4 m/s
The total weight increase by 42%
coefficient of rolling friction decrease by 19%
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
F(h) = F(f)
F(h) = mg* u
m is mass
g is gravitational acceleration = 9.8 m/s^2
200 = mg*u
Since weight increases by 42% and friction coefficient decreases by 19%
New weight = 1+0.42 = 1.42 = (1.42*m*g)
New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u
F(h) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
Answer:
6.13 s
219 N
Explanation:
Newton's law in the x direction:
∑F = ma
150 cos 30° N − 50 N = (30 kg) a
a = 2.66 m/s²
Δx = v₀ t + ½ at²
(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²
t = 6.13 s
Newton's law in the y direction:
∑F = ma
Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0
Fn = 219 N
Answer:
1900 meters
Explanation:
30m/s x 30 second = 900 meters
+ 1000 meters starting position
= 1900meters
The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is
∑ F = F[a] - F[f] - F[air] = ma
3100 N - 200 N - F[air] = (650 kg) (3 m/s²)
Solve for F[air] :
F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)
F[air] = 3100 N - 200 N - 1950 N
F[air] = 950 N
