First you need to make a difference between friction while object is stationary and the friction while object is moving. Force required to start moving some object is slightly greater than force required to maintain objects movement. That means that to move a chair you need some force F1 but you can than slightly reduce force and chair will still be moving.
Now to the problem in this question: It can be said that "stationary friction force" is equal to 15 Newtons. Its also good to know that friction force between chair and floor while you are increasing your push is also increasing and is equal to force of your push. Once it reaches 15N which is it "critical value" for that chair, chair starts moving and friction force drops a little bit and now it is called friction force of moving chair.
I got you
Explanation:
normal force = 400 g cos 35
friction force up slope = .6 (400 g) cos 35
weight component down slope = 400 g sin 35
400 a = 400 g sin 35 - .6 (400 g cos 35)
a = g (sin 35 - .6 cos 35) = .082 g
I hope this helps you
Answer:
i know the questin but i got to try and find it
Explanation:
Answer:
The speed with which the man flies forward is 5.5 m/s
Explanation:
The mass of the man = 100 kg
The mass of the scooter = 10 kg
The speed with which the man was traveling on the scooter = 5 m/s
The speed of the scooter after it hits the rock = 0 m/s
Let v represent the speed with which the man flies forward
The formula for momentum, P, is P = Mass × Velocity
The conservation of linear momentum principle is, the total initial momentum = The total final momentum, therefore, we have;
The total initial momentum = (100 kg + 10 kg) × 5 m/s = 550 kg·m/s
The total final momentum = 100 kg × v + 10 kg × 0 m/s = 100 kg × v
When the momentum is conserved, we have;
550 kg·m/s = 100 kg × v
∴ v = 550 kg·m/s/(100 kg) = 5.5 m/s.
The speed with which the man flies forward = v = 5.5 m/s
Answer:
g_x = 3.0 m / s^2
Explanation:
Given:
- Change in length of spring [email protected] = 22.6 cm
- Time taken for 11 oscillations t = 19.0 s
Find:
- The value of gravitational free fall g_x at plant X:
Solution:
- We will assume a simple harmonic motion of the mass for which Time is:
T = 2*pi*sqrt(k / m ) ...... 1
- Sum of forces in vertical direction @equilibrium is zero:
F_net = k*x - m*g_x = 0
(k / m) = (g_x / x) .... 2
- substitute Eq 2 into Eq 1:
2*pi / T = sqrt ( g_x / x )
g_x = (2*pi / T )^2 * x
- Evaluate g_x:
g_x = (2*pi / (19 / 11) )^2 * 0.226
g_x = 3.0 m / s^2
