91Ó°ÊÓ

1. Kinetic energy,$KE=40.0\text{J}$ .
2. Initial speed, $v=4.0\text{m/s}$.

After seeing the graph, we can say that distance is , and kinetic energy is . We can find the normal force by using a kinetic energy equation.

Formula:

1. Work done$W=F.X$
2. Kinetic energy,$KE=\frac{1}{2}m{v}^{\text{2}}$

From the work-energy theorem, the change in the kinetic energy will be the work doneby the system. That means,

$\begin{array}{rcl}\mathrm{Δ}KE& =& W\\ & =& Fx\\ & & \end{array}$

Here F is the force, and x is the displacement that took place.

The displacement from the graph can be observed as,

$x=2\text{m}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} 0=F×2 \\ F=20\text{N} \end{gathered}$$

We can say that force is along the x direction ofthe ramp.

Then,

$F_x=20\text{N}$

Fromthe figure below, we can say that normal force and force in y direction balance each other, thus,

$N=F_{\text{y}}$

From the given figure, we can write the equation, which is,

$\begin{array}{rcl}mg& =& F\\ & =& \sqrt{\left(F_{\text{x}}^{\text{2}}\right)+\left(F_{\text{y}}^{\text{2}}\right)}\\ & & \end{array}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} mg=\sqrt{F_{\text{x}}^{\text{2}}+N^{\text{2}}} \\ N=\sqrt{{\left(mg\right)}^2-F_{\text{x}}^{\text{2}}} \end{gathered}$$ (1)

We know that kinetic energy can be calculated as,

$$\begin{gathered} \text{KE}=\frac{1}{2}m{v}^{\text{2}} \\ m=\frac{2KE}{v^2} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} m=\frac{2\left(40\right)}{4^2} \\ m=5.0\text{kg} \end{gathered}$$

Substitute the values in equation 1, and we get,

$\begin{array}{rcl}N& =& \sqrt{{\left(5\text{kg}×9.8{\text{m/s}}^{\text{2}}\right)}^2-{\left(20\text{N}\right)}^2}\\ & =& \sqrt{{\left(49\left(1\text{kg}·{\text{m/s}}^{\text{2}}×\frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\right)\right)}^2-\left(20\text{N}\right)}\\ & =& \sqrt{{\left(49\text{N}\right)}^2-{\left(20\text{N}\right)}^2}\\ N& =& 44.73\text{N}\\ & ≈& 45\text{N}\\ & & \end{array}$

Thus,the normal force on the block is 45N