Question

A block of mass 10 kg slides down a rough
slope which is inclined at 45° to the horizontal.
The coefficient of sliding friction is 0.30. When
the block has to slide 5 m, the work done on
the block by the force of friction is nearly

Answer 1

Explanation:

4 votes

F = mue × mg So, F= 0.2×5×10 F=10 Now, W= mg cos (theta) W= 10× 10 × √2\2 So , W = 50 ... More

Answer 2

ANSWER:

The work done by the mass of block is 106.05 J

EXPLANATION:

The work done by the block of mass is the amount of the force applied on the body over a distance of “S”.

The sliding friction is shown by μ = 0.3

The distance of slide is 5 cm

The force applied on the body is shown as

$F=\mu m g \cos \theta$

$F=0.3 \times 10 \times 10 \times \cos 45$

$F=21.21 \mathrm{N}$

The work done formula is W = F.S

Now the work is done by force of friction therefore, loss in work or negative work meaning the work done is  

W = 21.21 x 5 = 106.05J

Therefore, the "work done" by the mass of block is 106.05J