Correct answer is (300)

To find the power delivered by the pulling force at t = 10 s, we first need to find the work done by the force. The work done is given by the product of force and displacement, and the power is the rate of work done.

Calculate the velocity (v) at t = 10 s:

Since the block starts from rest and is pulled up with an effective acceleration of 1 m/s², we can use the equation of motion to find the velocity (v) at t = 10 s:

$v=u+at$

Here, u = 0 (initial velocity) and a = 1 m/s² (acceleration). Plugging in the values:

$v_{10}=0+1(10)=10\mathrm{m}/\mathrm{s}$

1. Calculate the net force acting on the block ($F_{net}$):
   
   The net force acting on the block along the incline plane is the difference between the pulling force (F) and the gravitational force component acting parallel to the incline (mgsinθ):

$F_{\text{net}}=F-mgsin\theta$

Since F_net = ma, we can write:

$F=ma+mgsin\theta$

Plugging in the values (m = 5 kg, a = 1 m/s², g = 10 m/s², and θ = 30°):

$F=5(1)+5(10)(\sin 30°)=5+25=30\mathrm{N}$

Calculate the power (P) at t = 10 s:

The power (P) can be calculated as the product of force (F) and velocity (v):

$P_{10}=Fv=30(10)=300\mathrm{W}$

So, the power delivered by the pulling force at t = 10 s from the start is 300 W.

Correct answer is (32)

To find the work done by a force during a displacement, we can use the formula:

$W={\int }_{x_1}^{x_2}\overrightarrow{F}\cdot d\overrightarrow{x}$

Here, the force is given by $\overrightarrow{F}=(2+3x)\hat{i}$, and we need to find the work done during a displacement from $x=0$ to $x=4\mathrm{m}$. Since the force is only in the $x$ direction, we can write the integral as:

$W={\int }_0^4(2+3x)dx$

Now we can integrate the function with respect to $x$:

$W={\int }_0^4(2+3x)dx={\int }_0^{4}2dx+{\int }_0^{4}3xdx$

$W={\left[2x\right]}_0^4+{\left[\frac{3}{2}{x}^2\right]}_0^4$

Now we can plug in the limits of integration:

$W=(2\cdot 4-2\cdot 0)+(\frac{3}{2}\cdot 4^2-\frac{3}{2}\cdot 0^2)$

$W=8+24$

$W=32\mathrm{J}$

So the work done by the force during the displacement from $x=0$ to $x=4\mathrm{m}$ is 32 Joules.

Correct answer is (132)

$\begin{array}{rl}& W=\int Fdy={\int }_2^5(5+3y^2)dy\\ \\ & ={(5y+y^3)|}_2^5\\ \\ & =(15+125-8)\mathrm{J}\\ \\ & =132\mathrm{J}\end{array}$

Correct answer is (40)

The given expression calculates the work done by a force vector $\overrightarrow{F}=5\hat{i}+2\hat{j}+7\hat{k}$ when it acts on an object that moves from an initial position vector ${\overrightarrow{r}}_i=2\hat{i}+3\hat{j}-4\hat{k}$ to a final position vector ${\overrightarrow{r}}_f=5\hat{i}-2\hat{j}+\hat{k}$.

To find the work done, we use the dot product of the force and displacement vectors :

$\begin{array}{rl}& W=\overrightarrow{F}\cdot ({\overrightarrow{r}}_f-{\overrightarrow{r}}_{\mathrm{i}})\\ \\ & =(5\hat{i}+2\hat{j}+7\hat{k})\cdot ((5\hat{i}-2\hat{j}+\hat{k})-(2\hat{i}+3\hat{j}-4\hat{k}))\\ \\ & =(5\hat{i}+2\hat{j}+7\hat{k})\cdot (3\hat{i}-5\hat{j}+5\hat{k})\\ \\ & =15-10+35\\ \\ & =40\mathrm{J}\end{array}$

Correct Option is (A)

W_a = 0, W_b = +ve, W_c = +ve > W_b, W_d = $-$ve

$\Rightarrow$ W_c > W_b > W_a > W_d

$\Rightarrow$ W₃ > W₂ > W₁ > W₄