Question

a 20kg block is initially at rest .a 75n force is required to set the block in motion. after the motion starts, a force of 60n is applied to keep the block moving with constant speed.the coefficient of static friction is ??​

Answer 1

Answer:

${ \huge{ \underline{ \rm{ \large{ \pink{Given:}}}}}}$

• Force (F) = 75n
• Mass (m) = 20kg

${ \huge{ \underline{ \rm{ \large{ \green{Find:}}}}}}$

• Coefficient of static Friction?

${ \huge{ \underline{ \rm{ \large{ \red{Solution:}}}}}}$

For finding coefficient of static friction:-

• The coefficient of static friction is the friction force between two objects when neither of the objects is moving

• ${ \sf{μ= \frac{F}{mg} }}$

• So, by using this formula we can solve this.

$\to{ \sf{μ= \frac{F}{mg} }}$

${ \to{ \sf{ \frac{75}{20 \times 9.8} }}}$

${ \to{ \sf{ \frac{3.75}{9.8} = 0.38265}}}$

Therefore,

• The coefficient of static friction is 0.38265

Answer 2

Answer:

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