Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
<h3><u>Answer</u>;</h3>
= F0 L ( 1 - 1/e )
<h3><u>Explanation;</u></h3>
Work done is given as the product of force and distance.
In this case;
Work done = ∫︎ F(x) dx
= F0 ∫︎ e^(-x/L) dx
= F0 [ -L e^(-x/L) ] between 0 and L
= F0 L ( 1 - 1/e )
That would be called VOLT
Force can alter its direction,slow or stop it you could say it can change its velocity
