The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude <em>w</em>), the normal force (mag. <em>n</em>), and the friction force (maximum mag. <em>f</em> ).
In the horizontal direction, we have
<em>n</em> cos(120º) + <em>f</em> cos(30º) = 0
-1/2 <em>n</em> + √3/2 <em>f</em> = 0
<em>n</em> = √3 <em>f</em>
and in the vertical,
<em>n</em> sin(120º) + <em>f</em> sin(30º) + (-<em>w</em>) = 0
<em>n</em> sin(120º) + <em>f</em> sin(30º) = (50 kg) (9.80 m/s²)
√3/2 <em>n</em> + 1/2 <em>f</em> = 490 N
Substitute <em>n</em> = √3 <em>f</em> and solve for <em>f</em> :
√3/2 (√3 <em>f </em>) + 1/2 <em>f</em> = 490 N
2 <em>f</em> = 490 N
<em>f</em> = 245 N
(pointed up the incline)
1. 100,000% A. I simply googled C12 and it's the most stable Isotope of Carbon...
2. A as well.. 99% sure.. and so far my 99% sure has been 100% correct so ... :p
Answer:
(a) The work done is 0.05 J
(b) The force will stretch the spring by 3.8 cm
Explanation:
Given;
work done in stretching the spring from 30 cm to 45 cm, W = 3 J
extension of the spring, x = 45 cm - 30 cm = 15 cm = 0.15 m
The work done is given by;
W = ¹/₂kx²
where;
k is the force constant of the spring
k = 2W / x²
k = (2 x 3) / (0.15)²
k = 266.67 N/m
(a) the extension of the spring, x = 37 cm - 35 cm = 2 cm = 0.02 m
work done is given by;
W = ¹/₂kx²
W = ¹/₂ (266.67)(0.02)²
W = 0.05 J
(b) force = 10 N
natural length L = 30 cm
F = kx
x = F / k
x = 10 / 266.67
x = 0.0375 m
x = 3.75 cm = 3.8 cm
Thus a force of 10 N will stretch the spring by 3.8 cm
