
$300\;{\rm{J}}$ of work is done in sliding a $2\;{\rm{kg}}$ block up an inclined plane of height $10\;{\rm{m}}$. Work done against friction is (Take $g = 10\;{\rm{m}}/{{\rm{s}}^2}$)
(a) $1000\;{\rm{J}}$
(b) $200\;{\rm{J}}$
(c) $100\;{\rm{J}}$
(d) zero
Answer
205.5k+ views
Hint: The above problem is based on the work done under the effect of friction. The work done depends on the change in the position and direction of the force applied on the object. The loss of work occurs due to the presence of the friction.
Complete step by step answer
Given: The work done to slide the block is $W = 300\;{\rm{J}}$, mass of the block is $m = 2\;{\rm{kg}}$, height of the is $h = 10\;{\rm{m}}$, gravitational acceleration is $g = 10\;{\rm{m}}/{{\rm{s}}^2}$.
The formula to calculate the work done to raise the block on the inclined plane is given as:
${W_1} = mgh$
Substitute $2\;{\rm{kg}}$for m, $10\;{\rm{m}}/{{\rm{s}}^2}$for g and $10\;{\rm{m}}$for h in the above formula to find the work done to raise the block on the inclined plane.
${W_1} = \left( {2\;{\rm{kg}}} \right)\left( {10\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {10\;{\rm{m}}} \right)$
${W_1} = 200\;{\rm{J}}$
The formula to calculate the work done against friction is given as:
${W_f} = W - {W_1}$
Substitute $300\;{\rm{J}}$for $W$ and $200\;{\rm{J}}$for ${W_1}$ in the above expression to find the work done against friction.
${W_f} = 300\;{\rm{J}} - 200\;{\rm{J}}$
${W_f} = 100\;{\rm{J}}$
Thus, the work done against friction by the block is $100\;{\rm{J}}$ and the option (c) is the correct answer.
Additional Information: The work done by the object is divided into two types. First one is the work done by the conservative force and second one is the non-conservative force. The total mechanical energy of the system remains conserved if conservative forces are involved in the system. The change in the total energy remains conserved if non- conservative forces are involved in the system.
Note: The work done by the object is equal to the work done by the object against gravity and work done by the object against the friction. The work done by the object also equals the change in the kinetic energy of the object.
Complete step by step answer
Given: The work done to slide the block is $W = 300\;{\rm{J}}$, mass of the block is $m = 2\;{\rm{kg}}$, height of the is $h = 10\;{\rm{m}}$, gravitational acceleration is $g = 10\;{\rm{m}}/{{\rm{s}}^2}$.
The formula to calculate the work done to raise the block on the inclined plane is given as:
${W_1} = mgh$
Substitute $2\;{\rm{kg}}$for m, $10\;{\rm{m}}/{{\rm{s}}^2}$for g and $10\;{\rm{m}}$for h in the above formula to find the work done to raise the block on the inclined plane.
${W_1} = \left( {2\;{\rm{kg}}} \right)\left( {10\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {10\;{\rm{m}}} \right)$
${W_1} = 200\;{\rm{J}}$
The formula to calculate the work done against friction is given as:
${W_f} = W - {W_1}$
Substitute $300\;{\rm{J}}$for $W$ and $200\;{\rm{J}}$for ${W_1}$ in the above expression to find the work done against friction.
${W_f} = 300\;{\rm{J}} - 200\;{\rm{J}}$
${W_f} = 100\;{\rm{J}}$
Thus, the work done against friction by the block is $100\;{\rm{J}}$ and the option (c) is the correct answer.
Additional Information: The work done by the object is divided into two types. First one is the work done by the conservative force and second one is the non-conservative force. The total mechanical energy of the system remains conserved if conservative forces are involved in the system. The change in the total energy remains conserved if non- conservative forces are involved in the system.
Note: The work done by the object is equal to the work done by the object against gravity and work done by the object against the friction. The work done by the object also equals the change in the kinetic energy of the object.
Recently Updated Pages
JEE Mains Correction Window 2026- Session 1 and 2 Dates, Form Edit Link, Fee

JEE Main 2026 Marking Scheme- Marks Distribution, Negative and Total Marks

Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Trending doubts
Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Geostationary and Geosynchronous Satellites Explained

Charging and Discharging of Capacitor Explained

JEE Main 2025 Exam Pattern (Revised)

Photoelectric Effect and Stopping Potential: Concept, Formula & Exam Guide

JEE Main 2026 Session 1 Application Form Opening Soon – Important Dates & Details

Other Pages
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 10 Thermal Properties of Matter - 2025-26

JEE Advanced 2026 Revision Notes for Practical Organic Chemistry

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

