The elastic potential energy of a spring is given by
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
In our problem, initially the spring is uncompressed, so
. Therefore, the work done by the spring when it is compressed until
is
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
Answer:
Net force exerted on the radio is 27.5 Newton.
Given:
Mass = 5.5 kg
Acceleration = 5 
To find:
Force exerted on the radio = ?
Formula used:
F = ma
Where F = net force
m = mass
a = acceleration
Solution:
According to Newton's second law of motion,
F = ma
Where F = net force
m = mass
a = acceleration
F = 5.5 × 5
F = 27.5 Newton
Hence, Net force exerted on the radio is 27.5 Newton.
Answer:
resultant force = (f1²+f2²)½
=(1.5²+2²)½
=(2.25+4)½
=(6.25)½
=2.5
Explanation:
okay this question seems easy. now if the 1.5 is vertically upwards so is that 2 is horizontally downwards hence as u say its 90 degrees thn it forms a right angled triangle.
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
