Question

A block is sent up a friction less ramp along which an $x$ axis extends upward. From figure, it shows the kinetic energy of the block as a function of position $x$; the scale of the figure’s vertical axis is set by $K_s=40.0J$. If the block’s initial speed is $4.00m/s$, what is the normal force on the block?

Answer 1

From the figure, one may write the kinetic energy (in units of J) as a function of $x$ as

$K=K_s-20x=40-20x$.

Since $W=\Delta K={\overrightarrow{F}}_x.\Delta x$,

the component of the force along the force along $+x$ is

$F_x=dK/dx=-20N$.

The normal force on the block is $F_N=F_y$, which is related to the gravitational force by

$mg=\sqrt{F_x^2+(-F_y{)}^2}$.

$F_N$ points in the opposite direction of the component of the gravitational force.

With an initial kinetic energy

$K_s=40.0J$ and $v_0=4.00m/s$, the mass of the block is

$m=\frac{2K_s}{v_0^2}=\frac{2\left(40.0J\right)}{(4.00m/s{)}^2}=5.00kg$.

Thus, the normal force is$F_y=\sqrt{(mg{)}^2-F_x^2}$

$=\sqrt{\left(5.0kg{)}^2\right(9.8m/s^2{)}^2-(20.N{)}^2}=44.7N\approx 45N$.