Simple Problem #04

Classical Mechanics Level 3

A block slides on a plane with $45^{\circ}$ inclination without friction.

As shown, there are two sensors on this plane, $S_1$ and $S_2$ , which are not drawn to scale. When the block passed $S_1,$ it registered the time $t_1 = 3 \text{ s}$ , and when the block passed $S_2,$ it registered $t_2 = 5 \text{ s}$ .

Given that the initial velocity of the block was zero, the acceleration of gravity $g = 10 \text{ m/s}^2,$ and $\sqrt{2} = 1.41$ , determine the distance (in meters) between the two sensors.

The answer is 56.4.

1 solution

Vijay Simha
Feb 8, 2018

S1 = ut + 0.5 g sin(45)(t0)^2,

Since u = 0,

S1 = 0.5 g sin(45)*(3)^2

S2 = ut + 0.5 g sin(45)(t1)^2

Again, Since u = 0,

S2 = 0.5 g sin(45)(5)^2

S2 - S1 = 0.5 g sin(45) (25-9) = 0.5 g sin(45) 16 = (0.5X10X16)/(sqrt(2)) = 80/sqrt(2) = 56.56 m

You can also rewrite this solution in LaTeX :

$S_{1}=u \times t + 0.5 \times g \times \sin{45 ^\circ} \times (t_{0})^{2}$ .

Since $u=0$ ,

$S_{1} = 0.5 \times g \times \sin{45 ^\circ} \times 3^{2}$ .

$S_{2}=u \times t + 0.5 \times g \times \sin{45 ^\circ} \times (t_{1})^{2}$ ,

Again, Since $u=0$ ,

$S_{2} = 0.5 \times g \times \sin{45 ^\circ} \times 5^{2}$ .

$S_{2} - S_{1} = 0.5 \times g \times \sin{45 ^\circ} \times (25-9) = 0.5 \times g \times \sin{45 ^\circ} \times 16$ ,

$S_{2} - S_{1} = \frac{0.5 \times 10 \times 16}{\sqrt{2}} = \frac{80}{\sqrt{2}} = 40 \times \sqrt{2}$ ,

$S_{2} - S_{1} \approx 40 \times 1.41 = \boxed{56.4}$ meters.

Filip Rázek - 3 years, 3 months ago

Yup. Pretty much what I did. I used equation for acceleration along a slope by newton's second law, applied it to the kinematic equation of motion for displacement, and subtracted the two displacements.

Krishna Karthik - 2 years, 2 months ago

Simple Problem #04

The answer is 56.4.

1 solution

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