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3 years ago

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A box with a mass of 7 kilograms is pushed up a ramp to a height of 5 meters. What is the work done against the force of gravity

? Assume the ramp has negligible friction.
A. 3.4 × 102 joules

B. 6.9 × 102 joules

C. 8.6 × 102 joules

D. 1.7 × 102 joules

2 answers:

zloy xaker [14]3 years ago

7 0

A. 3.4 x 10 2 its right i guessed it

Musya8 [376]3 years ago

4 0

The correct answer to the question is $3.4\times 10^2\ J$ .

CALCULATION:

As per the question, the mass of the box is given as m = 7 kg.

The box was pushed to a height h = 5 m.

We are asked to calculate the work done by the box.

We know that energy is the capacity to do work.The work done by a body when it is taken to certain height above the surface of earth against gravity is equal to the potential energy gained by that body .

Hence, the work done W = P.E

The potential energy gained by the box when it was taken to certain height is the gravitational potential energy of the box.

The gravitational potential energy gained by the box is calculated as -

Potential energy P.E = mgh

= 7 × 9.8 × 5 J.

= 343 J

= $3.43\times 10^2\ J$

= $3.4\times 10^2\ J$

Here, g is known as the acceleration due to gravity and g = $9.8\ m/s^2$

We know that work done by the box against gravity W = P.E.

Hence, the work done by the box against gravity is $3.4\times 10^2\ J$

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Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

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3 years ago