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# Five persons $A, B, C, D$ and $E$, are pulling a cart of mass 100kg on a smooth surface and the cart is moving with acceleration $3\,m/{s^2}$ in the east direction. When person A stops pulling, it moves with acceleration $24\,m/{s^2}$ in the north direction. The magnitude of acceleration of the cart when only $A$ and $B$ pull the cart keeping their direction, is:A. $26\,m/{s^2}$B. $3\sqrt {71}\,m/{s^2}$C. $25\,m/{s^2}$D. $30\,m/{s^2}$

Last updated date: 25th Jul 2024
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Hint: Newton's laws of motion are three laws that explain the relationship between an object's motion and the forces acting on it in classical mechanics. Until it is acted upon by an external force, the first law states that an object either stays at rest or begins to travel at a constant velocity.

$F = ma$
$m = 100\,kg$
From the given data we obtain
${F_a} + {F_b} + {F_c} + {F_d} + {F_e} = 100.3i$ - (1)
$i$ being the unit vector in the east direction.
Also we obtain
${F_b} + {F_c} + {F_d} + {F_e} = 100. - 1i$ - (2)
And
${F_a} + {F_c} + {F_d} + {F_e} = 100.24j$ - (3)
From 1 and 2
${F_a} = 100.4i$
From 1 and 3
${F_b} = 100.(3i - 24j)$
Suppose if A and B only are pulling
${F_a} + {F_b} = 100.(7i - 24j)$
here
$\vec a = \dfrac{{{{\vec F}_A} + {{\vec F}_B}}}{m} = \dfrac{{(700\hat i - 2400\hat j)}}{{100}}m/{s^2}$
$\Rightarrow \vec a = (7\hat{i} - 24\hat {j})\,m/{s^2}$
$\therefore |\vec a| = \sqrt {{7^2} + {{24}^2}} = 25\,m/{s^2}$

Hence, the magnitude of acceleration of the cart is $25\,m/{s^2}$.

Note: The rate of change of an object's velocity with respect to time is called acceleration in mechanics. Accelerations are quantities that are measured in vectors (in that they have magnitude and direction). The orientation of the net force acting on an object determines the orientation of its acceleration.