Question

A block of mass m is placed on a smooth wedge of inclination

Answer 1

Answer:

Explanation:

For an observer on the ground

Rsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg

⇒a=gtanθ⇒a=gtanθ

(ii) F=(M+m)a=(M+m)gtanF=(M+m)a=(M+m)gtan●

iii) Force exerted by the wedge on the block

⇒=mgcosθorR=mgsecθ⇒=mgcosθorR=mgsecθ

If inclination is given as 1 in x, sinθ=1xsinθ=1x

tan=1x2−1−−−−−√tan=1x2-1

⇒⇒ Acceleration a=gtan=gx2−1−−−−−√

Answer 2

For an observer on the ground

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg⇒a=gtanθ⇒a=gtanθ

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg⇒a=gtanθ⇒a=gtanθ(ii) F=(M+m)a=(M+m)gtanF=(M+m)a=(M+m)gtan●

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg⇒a=gtanθ⇒a=gtanθ(ii) F=(M+m)a=(M+m)gtanF=(M+m)a=(M+m)gtan●iii) Force exerted by the wedge on the block

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg⇒a=gtanθ⇒a=gtanθ(ii) F=(M+m)a=(M+m)gtanF=(M+m)a=(M+m)gtan●iii) Force exerted by the wedge on the block⇒=mgcosθorR=mgsecθ⇒=mgcosθorR=mgsecθ

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg⇒a=gtanθ⇒a=gtanθ(ii) F=(M+m)a=(M+m)gtanF=(M+m)a=(M+m)gtan●iii) Force exerted by the wedge on the block⇒=mgcosθorR=mgsecθ⇒=mgcosθorR=mgsecθIf inclination is given as 1 in x, sinθ=1xsinθ=1x

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg⇒a=gtanθ⇒a=gtanθ(ii) F=(M+m)a=(M+m)gtanF=(M+m)a=(M+m)gtan●iii) Force exerted by the wedge on the block⇒=mgcosθorR=mgsecθ⇒=mgcosθorR=mgsecθIf inclination is given as 1 in x, sinθ=1xsinθ=1xtan=1x2−1−−−−−√tan=1x2-1

For an observer on the groundRsinθ=ma,Rcosθ=mgRsinθ=ma,Rcosθ=mg⇒a=gtanθ⇒a=gtanθ(ii) F=(M+m)a=(M+m)gtanF=(M+m)a=(M+m)gtan●iii) Force exerted by the wedge on the block⇒=mgcosθorR=mgsecθ⇒=mgcosθorR=mgsecθIf inclination is given as 1 in x, sinθ=1xsinθ=1xtan=1x2−1−−−−−√tan=1x2-1⇒⇒ Acceleration a=gtan=gx2−1−−−−−√