Question

Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy plane. One force is $\vec{F}_1 = (3.9 N)\hat{i} + (3.3 N)\hat{j}$. Find the acceleration of the chopping block in unit-vector notation for each of the following second forces.$a) \vec{F}_2= (-3.0N)\hat{i} + (-4.0N)\hat{j}\\b) \vec{F}_2= (-3.0N)\hat{i} + (4.0N)\hat{j}\\c) \vec{F}_2=(3.0N)\hat{i} + (-4.0N)\hat{j}$

Answer 1

Answer:

Part a)

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

Explanation:

As per Newton's II law we know that

F = ma

so we will have

$a = \frac{F}{m}$

so we will have

$a = \frac{F_1 + F_2}{m}$

Part a)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}$

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}$

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}$

$a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}$

$a = (4.92\hat i - 0.5 \hat j)m/s^2$