dripcima24
2022-10-02
Answered

Two forces act on a 2.55 kg object, the gravitational force and a second, constant force. The object starts from rest and in 1.20 s is displaced $(4.05\hat{i}-3.30\hat{j})$ m. Write second force in unit vector notation.

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Iademarco11

Answered 2022-10-03
Author has **10** answers

Given:

m =2.55 Kg

t =1.20 Sec

$d=4.05\hat{i}-3.30\hat{j}$

To find:

Force acting on the given mass.

Solution:

Gravitational force = ${m}_{g}=2.55\times 9.8=24.99\text{}N$

Direction would be - $\hat{j}$

Constant force acting on

As gravitational force acting in downward direction constant force should be in positive x direction as shown in displacement.

Only x direction displacement is to be taken here.

$\text{velocity}=\frac{\text{displacement}}{\text{time}}\phantom{\rule{0ex}{0ex}}\text{velocity}=\frac{4.05\hat{i}}{1.20}=3.375\hat{i}\phantom{\rule{0ex}{0ex}}\text{acceleration}=\frac{\text{velocity}}{\text{time}}\phantom{\rule{0ex}{0ex}}\text{acceleration}=\frac{3.375}{1.20}\hat{i}=2.81\hat{i}$

Force is the product of mass and acceleration.

$\text{force}=\text{mass}\times \text{acceleration}\phantom{\rule{0ex}{0ex}}\text{force}=2.55\times 2.81\hat{i}\phantom{\rule{0ex}{0ex}}\text{force}=7.16\text{}kg\text{}m/{s}^{2}$

direction is positive x direction

$\text{force}=7.16\text{}kg\text{}m/{s}^{2}$

m =2.55 Kg

t =1.20 Sec

$d=4.05\hat{i}-3.30\hat{j}$

To find:

Force acting on the given mass.

Solution:

Gravitational force = ${m}_{g}=2.55\times 9.8=24.99\text{}N$

Direction would be - $\hat{j}$

Constant force acting on

As gravitational force acting in downward direction constant force should be in positive x direction as shown in displacement.

Only x direction displacement is to be taken here.

$\text{velocity}=\frac{\text{displacement}}{\text{time}}\phantom{\rule{0ex}{0ex}}\text{velocity}=\frac{4.05\hat{i}}{1.20}=3.375\hat{i}\phantom{\rule{0ex}{0ex}}\text{acceleration}=\frac{\text{velocity}}{\text{time}}\phantom{\rule{0ex}{0ex}}\text{acceleration}=\frac{3.375}{1.20}\hat{i}=2.81\hat{i}$

Force is the product of mass and acceleration.

$\text{force}=\text{mass}\times \text{acceleration}\phantom{\rule{0ex}{0ex}}\text{force}=2.55\times 2.81\hat{i}\phantom{\rule{0ex}{0ex}}\text{force}=7.16\text{}kg\text{}m/{s}^{2}$

direction is positive x direction

$\text{force}=7.16\text{}kg\text{}m/{s}^{2}$

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