
A block of mass 10 kg is kept on a horizontal surface. A force F is acted on the block as shown in figure. For what minimum value of F, the block will be lifted up?
A.98 N
B.49 N
C.200 N
D.None of these
Answer
582.9k+ views
Hint: For the block of mass 10 kg to be lifted up the force lifting it up should be greater than the force pulling down or the force lifting it up should be equal to the force pulling down. So, first find the force lifting the block i.e. upward force and then find the force pulling the block down i.e. downward force. Substitute these forces in the above-mentioned condition and find the minimum force required to lift the block.
Complete answer:
Given: Mass of the block (m)= 10 kg
$\theta = 60°$
Upward force is given by,
${F}_{up}= F \cos {\theta}$
Substituting the value in above equation we get,
${F}_{up}= F \cos {60°}$
$\Rightarrow {F}_{up}= \dfrac {F}{2}$
Downward force is given by,
${F}_{down}= mg$
Substituting the values in above equation we get,
${F}_{down}= 10 \times 10$
$\Rightarrow {F}_{down}= 100 N$
For the block to be lifted by the upward should be either greater than or equal to the downward force.
$\therefore {F}_{up} \geq {F}_{down}$
Substituting the values we get,
$\dfrac {F}{2} \geq 100$
$\Rightarrow F \geq 2 \times 100$
$\Rightarrow F \geq 200 N$
Hence, force will be lifted up to a minimum of 200N.
So, the correct answer is option C i.e. 200 N.
Note:
To solve this problem, the first step is to make a free body diagram to resolve the forces. Making a force body diagram makes it easier for us to understand the unknown forces and get the clear idea of all the forces acting on the body. Applying force on a body can also change the direction of the object. It also changes the shape and size of the object. Applying force on an object can either stop the moving object or a stationary object. It can also increase or decrease the speed of the object.
Complete answer:
Given: Mass of the block (m)= 10 kg
$\theta = 60°$
Upward force is given by,
${F}_{up}= F \cos {\theta}$
Substituting the value in above equation we get,
${F}_{up}= F \cos {60°}$
$\Rightarrow {F}_{up}= \dfrac {F}{2}$
Downward force is given by,
${F}_{down}= mg$
Substituting the values in above equation we get,
${F}_{down}= 10 \times 10$
$\Rightarrow {F}_{down}= 100 N$
For the block to be lifted by the upward should be either greater than or equal to the downward force.
$\therefore {F}_{up} \geq {F}_{down}$
Substituting the values we get,
$\dfrac {F}{2} \geq 100$
$\Rightarrow F \geq 2 \times 100$
$\Rightarrow F \geq 200 N$
Hence, force will be lifted up to a minimum of 200N.
So, the correct answer is option C i.e. 200 N.
Note:
To solve this problem, the first step is to make a free body diagram to resolve the forces. Making a force body diagram makes it easier for us to understand the unknown forces and get the clear idea of all the forces acting on the body. Applying force on a body can also change the direction of the object. It also changes the shape and size of the object. Applying force on an object can either stop the moving object or a stationary object. It can also increase or decrease the speed of the object.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

